NCERT Class 11 Maths Chapter 4 or Exercise 4.1 Solutions – Principle of Mathematical Induction

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Prove the following by using the principle of mathematical induction for all n ∈ N

1. 1 + 3 + 32 + ……….. + 3n-1 = (3n – 1)/2

Let’s suppose that P(n) : 1 + 3 + 32 + ……….. + 3n-1 = (3n – 1)/2

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1
Right Hand Side of P(n) = (3 × 1 – 1)/2 = (3 – 1)/2 = 2/2 = 1

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 1

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1 + 3 + 32 + ……….. + 3k-1 = (3k – 1)/2 is True [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : 1 + 3 + 32 + ……….. + 3(k + 1) – 1 = (3k + 1 – 1)/2
Simplifying this
⇒ P(k + 1) : 1 + 3 + 32 + ……….. + 3k = (3k + 1 – 1)/2 (We need to prove this) [Equation 2]

Using [Equation 1]
⇒ P(k) : 1 + 3 + 32 + ……….. + 3k-1 = (3k – 1)/2
Adding 3k to both sides of above equation
So above equation can be written as
1 + 3 + 32 + ……….. + 3k-1 + 3k = (3k – 1)/2 + 3k

Let’s now simplify Right Hand Side of above equation
1 + 3 + 32 + ……….. + 3k-1 + 3k = (3k – 1 + 2 × 3k)/2 = (3 × 3k – 1)/2 = (3k + 1 – 1)/2

⇒ 1 + 3 + 32 + ……….. + 3k-1 + 3k = (3k + 1 – 1)/2 which is same as [Equation 2] which we need to prove True

Thus P(n) is True for n = k

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



2. 13 + 23 + 33 + ………… + n3 = (n(n + 1)/2)2

Let’s suppose that P(n) : 13 + 23 + 33 + ………… + n3 = (n(n + 1)/2)2

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 13 = 1
Right Hand Side of P(n) = (1(1 + 1)/2)2 = (2/2)2 = 12 = 1

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 1

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 13 + 23 + 33 + ………… + k3 = (k(k + 1)/2)2 is True [Equation 1]

Now let’s prove that P(k + 1) is also True

cubic equation

Using [Equation 1]
⇒ P(k) : 13 + 23 + 33 + ………… + k3 = (k(k + 1)/2)2
Adding (k + 1)3 to both sides of this equation

⇒ P(k) : 13 + 23 + 33 + ………… + k3 + (k + 1)3 = (k(k + 1)/2)2 + (k + 1)3
Let’s simplify this

Principle of Mathematical Induction for Cubic Equation

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


3. 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ………….. + 1/(1 + 2 + 3 + …….. + n) = 2n/(n + 1)

Let’s suppose that P(n) : 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ………….. + 1/(1 + 2 + 3 + …….. + n) = 2n/(n + 1)

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1/1 = 1
Right Hand Side of P(n) = 2 × 1/(1 + 1) = 2/2 = 1

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 1

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(n) : 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ………….. + 1/(1 + 2 + 3 + …….. + k) = 2k/(k + 1) [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ………….. + 1/(1 + 2 + 3 + …….. + (k + 1)) = 2(k + 1)/((k + 1) + 1)
Simplify Right Hand Side of this equation
⇒ P(k + 1) : 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ………….. + 1/(1 + 2 + 3 + …….. + (k + 1)) = 2(k + 1)/(k + 2) (we need to prove this)

Left Hand side of this can be rewritten as

Thus as Left Hand is equals to Right Hand Side
⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



4. 1.2.3 + 2.3.4 + …………. + n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)/4

Let’s suppose that P(n) : 1.2.3 + 2.3.4 + …………. + n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)/4

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1(1 + 1)(1 + 2) = 1(2)(3) = 6
Right Hand Side of P(n) = 1(1 + 1)(1 + 2)(1 + 3)/4 = 1 × 2 × 3 × 4/4 = 1 × 2 × 3 = 6

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 6

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1.2.3 + 2.3.4 + …………. + k(k + 1)(k + 2) = k(k + 1)(k + 2)(k + 3)/4 [Equation 1]

Now let’s prove that P(k + 1) is also True

Solving Left Hand Side of above equation and proving it equals to Right Hand Side
Left Hand Side of above equation can be rewritten as following: –

Which is equals to Right Hand Side.

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


5. 1.3 + 2.32 + 3.33 + ………….. + n.3n = [(2n – 1)3n+1 + 3]/4

Let’s suppose that P(n) : 1.3 + 2.32 + 3.33 + ………….. + n.3n = [(2n – 1)3n+1 + 3]/4

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1.31 = 3
Right Hand Side of P(n) = [(2 × 1 – 1)31 + 1 + 3]/4 = [(2 – 1)32 + 3]/4 = [9 + 3]/4 = 12/4 = 3

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 3

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1.3 + 2.32 + 3.33 + ………….. + k.3k = [(2k – 1)3k+1 + 3]/4 [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : 1.3 + 2.32 + 3.33 + ………….. + (k + 1).3(k + 1) = [(2(k + 1) – 1)3(k + 1)+1 + 3]/4
Simplifying Right Hand Side of this equation

Left Hand Side of this equation can be rewritten as
1.3 + 2.32 + 3.33 + ………….. + k.3k + (k + 1).3(k + 1)

Using [Equation 1] replacing the value
⇒ [(2k – 1)3k+1 + 3]/4 + (k + 1).3(k + 1)

Simplifying this

\begin{equation} \frac{(2k – 1)3^{k + 1} + 3}{4} + (k + 1).3^{(k + 1)} \\ \frac{(2k – 1)3^{k + 1} + 3 + 4(k + 1).3^{(k + 1)}}{4} \\ \frac{(2k – 1)3^{k + 1} + 4(k + 1).3^{(k + 1)} + 3}{4} \\ \frac{3^{k + 1}[2k – 1 + 4(k + 1)] + 3}{4} \\ \frac{3^{k + 1}[2k – 1 + 4k + 4] + 3}{4} \\ \frac{(6k + 3)3^{k + 1} + 3}{4} \\ \frac{3(2k + 1)3^{k + 1} + 3}{4} \\ \frac{(2k + 1)3^{k + 2} + 3}{4} \\ \end{equation}

Which is what we need to prove.
Hence it’s proved that
P(k + 1) : 1.3 + 2.32 + 3.33 + ………….. + (k + 1).3(k + 1) = [(2(k + 1) – 1)3(k + 1)+1 + 3]/4

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



6. 1.2 + 2.3 + 3.4 + ……… + n.(n + 1) = [n(n + 1)(n + 2)/3]

Let’s suppose that P(n) : 1.2 + 2.3 + 3.4 + ……… + n.(n + 1) = [n(n + 1)(n + 2)/3]

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1.(1 + 1) = 1.2 = 2
Right Hand Side of P(n) = [1(1 + 1)(1 + 2)/3] = 2 × 3 /3 = 2

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 2

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1.2 + 2.3 + 3.4 + ……… + k.(k + 1) = [k(k + 1)(k + 2)/3] [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : 1.2 + 2.3 + 3.4 + ……… + (k + 1).((k + 1) + 1) = [(k + 1)(k + 1 + 1)(k + 1 + 2)/3]
Simplifying this equation
P(k + 1) : 1.2 + 2.3 + 3.4 + ……… + (k + 1).(k + 2) = [(k + 1)(k + 2)(k + 3)/3]
We need to prove that this equation is True

Left Hand Side of this equation can be rewritten as
1.2 + 2.3 + 3.4 + ……… + k.(k + 1) + (k + 1).(k + 2)

Using [Equation 1] to replace value of 1.2 + 2.3 + 3.4 + ……… + k.(k + 1)
[k(k + 1)(k + 2)/3] + (k + 1).(k + 2)

\begin{equation} \frac{k(k + 1)(k + 2)}{3} + (k + 1)\cdot(k + 2) \\ \frac{ (k + 1)\cdot(k + 2) \left[ \frac{k}{3} + 1 \right] }{4} \\ \frac{(k + 1)(k + 2)(k + 3)}{4} \\ \text{which is equals to Right Hand Side of P(k + 1)} \end{equation}

Hence it’s prove that P(k + 1) : 1.2 + 2.3 + 3.4 + ……… + (k + 1).(k + 2) = [(k + 1)(k + 2)(k + 3)/3] is also True.

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


7. 1.3 + 3.5 + 5.7 + ……… + (2n – 1)(2n + 1) = n(4n2 + 6n – 1)/3

Let’s suppose that P(n) : 1.3 + 3.5 + 5.7 + ……… + (2n – 1)(2n + 1) = n(4n2 + 6n – 1)/3

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = (2 × 1 – 1)(2 × 1 + 1) = (2 – 1)(2 + 1) = 1 × 3 = 3
Right Hand Side of P(n) = 1(4 × 12 + 6 × 1 – 1)/3 = (4 + 6 – 1)/3 = 9/3 = 3

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 3

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1.3 + 3.5 + 5.7 + ……… + (2k – 1)(2k + 1) = k(4k2 + 6k – 1)/3 [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2(k + 1) – 1)(2(k + 1) + 1) = (k + 1)(4(k + 1)2 + 6(k + 1) – 1)/3

Let’s simplify this
⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 2 – 1)(2k + 2 + 1) = (k + 1)(4(k2 + 2k + 1) + 6k + 6 – 1)/3
⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (k + 1)(4k2 + 8k + 4 + 6k + 6 – 1)/3

⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (k + 1)(4k2 + 14k + 9)/3
⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (4k3 + 14k2 + 9k + 4k2 + 14k + 9)/3

⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (4k3 + 18k2 + 23k + 9)/3
We need to prove that this equation is True

Rewriting Left Hand Side of this equation
1.3 + 3.5 + 5.7 + ……… + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)

Replacing value using [Equation 1]
⇒ k(4k2 + 6k – 1)/3 + (2k + 1)(2k + 3)

\begin{equation} \frac{k(4k^{2} + 6k – 1)}{3} + (2k + 1)(2k + 3) \\ \frac{k(4k^{2} + 6k – 1)}{3} + 2k(2k + 3) + 1(2k + 3) \\ \frac{k(4k^{2} + 6k – 1)}{3} + 4k^{2} + 6k + 2k + 3 \\ \frac{k(4k^{2} + 6k – 1)}{3} + 4k^{2} + 8k + 3 \\ \frac{k(4k^{2} + 6k – 1) + 3(4k^{2} + 8k + 3)}{3} \\ \frac{4k^{3} + 6k^{2} – k + 12k^{2} + 24k + 9}{3} \\ \frac{4k^{3} + 18k^{2} + 23k + 9}{3} \\ \text{which is same as Right Hand Side of P(k + 1)} \end{equation}

Hence its proved that if P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


8. 1.2 + 2.22 + 3.23 + ………. + n.2n = (n – 1)2n + 1 + 2

Let’s suppose that P(n) : 1.2 + 2.22 + 3.23 + ………. + n.2n = (n – 1)2n + 1 + 2

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1.21 = 2
Right Hand Side of P(n) = (1 – 1)21 + 1 + 2 = (0)22 + 2 = 2

⇒ Left Hand Side of P(n) = Right Hand Side of P(n) = 2

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1.2 + 2.22 + 3.23 + ………. + k.2k = (k – 1)2k + 1 + 2 [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : 1.2 + 2.22 + 3.23 + ………. + (k + 1).2k + 1 = ((k + 1) – 1)2(k + 1) + 1 + 2

Simplifying this equation
⇒ P(k + 1) : 1.2 + 2.22 + 3.23 + ………. + (k + 1).2k + 1 = (k + 1 – 1)2k + 1 + 1 + 2

⇒ P(k + 1) : 1.2 + 2.22 + 3.23 + ………. + (k + 1).2k + 1 = k.2k+2 + 2
We need to prove this

Let’s rewrite Left Hand Side of this equation
⇒ 1.2 + 2.22 + 3.23 + ………. + k.2k + (k + 1).2k + 1

Using [Equation 1] to rewrite this equation
⇒ (k – 1)2k + 1 + 2 + (k + 1).2k + 1

⇒ 2k + 1(k – 1 + k + 1) + 2

⇒ 2k + 1(2k) + 2

⇒ k.2k+2 + 2 which is equals to Right Hand Side of P(k + 1)

Thus if P(k) is True then P(k + 1) will also be True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



9. 1/2 + 1/4 + 1/8 + …………. + 1/2n = 1 – 1/2n

Let’s suppose that P(n) : 1/2 + 1/4 + 1/8 + …………. + 1/2n = 1 – 1/2n

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1/21 = 1/2
Right Hand Side of P(n) = 1 – 1/21 = 1 – 1/2 = 1/2

Left Hand Side of P(n) = Right Hand Side of P(n) = 1/2

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1/2 + 1/4 + 1/8 + …………. + 1/2k = 1 – 1/2k [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : 1/2 + 1/4 + 1/8 + …………. + 1/2k + 1 = 1 – 1/2k + 1

Let’s simplify Left Hand Side of P(k + 1)

\begin{equation} \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ ……… + \frac{1}{2^{k + 1}} \\ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ ……… + \frac{1}{2^{k}} + \frac{1}{2^{k + 1}} \\ \end{equation}

Using [Equation 1] to replace value in the above equation

\begin{equation} 1 – \frac{1}{2^{k}} + \frac{1}{2^{k + 1}} \\ 1 – \frac{1}{2^{k}}(1 – \frac{1}{2}) \\ 1 – \frac{1}{2^{k}}(\frac{2 – 1}{2}) \\ 1 – \frac{1}{2^{k}}(\frac{1}{2}) \\ 1 – \frac{1}{2^{k + 1}} \end{equation}

which is equals to Right Hand Side of P(k + 1)

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1/(3 × 1 – 1)(3 × 1 + 2) = 1/(3 – 1)(3 + 2) = 1/2 × 5 = 1/10

Right Hand Side of P(n) = 1/(6 × 1 + 4) = 1/(6 + 4) = 1/10

Left Hand Side of P(n) = Right Hand Side of P(n) = 1/10

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s start with Left Hand Side of this equation and simplify it

which is equals to Right Hand Side of P(k + 1)

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1/1(1 + 1)(1 + 2) = 1/(1)(2)(3) = 1/6
Right Hand Side of P(n) = 1(1 + 3)/4(1 + 1)(1 + 2) = 4/4(2)(3) = 1/(2)(3) = 1/6

Left Hand Side of P(n) = Right Hand Side of P(n) = 1/6

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Using [Equation 1] to replace value in this equation.

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


12. a + ar + ar2 + ………. + arn-1 = a(rn – 1)/r – 1

Let’s suppose that P(n) : a + ar + ar2 + ………. + arn-1 = a(rn – 1)/r – 1

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = ar1-1 = ar0 = a(1) = a
Right Hand Side of P(n) = a(r1 – 1)/r – 1 = a(r – 1)/r – 1 = a

Left Hand Side of P(n) = Right Hand Side of P(n) = a

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : a + ar + ar2 + ………. + ark-1 = a(rk – 1)/(r – 1) [Equation 1]

Now let’s prove that P(k + 1) is also True
⇒ P(k + 1) : a + ar + ar2 + ………. + ar(k + 1)-1 = a(r(k + 1) – 1)/(r – 1)

Let’s simplify this equation
⇒ P(k + 1) : a + ar + ar2 + ………. + ark = a(r(k + 1) – 1)/(r – 1)
We need to prove that this equation is True

Let’s simplify Left Hand Side of this equation
⇒ a + ar + ar2 + ………. + ark-1 + ark

Replacing value in this equation using [Equation 1]
⇒ a(rk – 1)/(r – 1) + ark

\begin{equation} \implies \frac{a(r^{k} – 1)}{(r – 1)} + ar^{k} \\ \implies \frac{a(r^{k} – 1) + ar^{k}(r – 1)}{(r – 1)} \\ \implies \end{equation}
\begin{equation} \frac{ar^{k} – a + ar^{k + 1} – ar^{k}}{(r – 1)} \\ \end{equation}
\begin{equation} \implies \frac{ar^{k} – a + ar^{k + 1} – ar^{k}}{(r – 1)} \\ \implies \frac{ar^{k + 1} – a}{(r – 1)} \\ \implies \frac{a(r^{k + 1} – 1)}{(r – 1)} \\ \text{Which is same as Right Hand Side of P(k + 1)} \end{equation}

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


13. (1 + 3/1)(1 + 5/4)(1 + 7/9)…………..(1 + (2n + 1)/n2) = (n + 1)2

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = (1 + (2 × 1 + 1)/12) = 1 + (2 + 1)/1 = 1 + 3 = 4
Right Hand Side of P(n) = (1 + 1)2 = 22 = 4

Left Hand Side of P(n) = Right Hand Side of P(n) = 4

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


14. (1 + 1/1)(1 + 1/2)(1 + 1/3) ………. (1 + 1/n) = (n + 1)

Let’s suppose that P(n) : (1 + 1/1)(1 + 1/2)(1 + 1/3) ………….. (1 + 1/n) = (n + 1)

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1 + 1/1 = 1 + 1 = 2
Right Hand Side of P(n) = 1 + 1 = 2

Left Hand Side of P(n) = Right Hand Side of P(n) = 2

Thus P(m) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ p(k) : (1 + 1/1)(1 + 1/2)(1 + 1/3) ………….. (1 + 1/k) = (k + 1) [Equation 1]

We need to prove that P(k + 1) is also True
⇒ p(k + 1) : (1 + 1/1)(1 + 1/2)(1 + 1/3) ………….. (1 + 1/(k + 1)) = ((k + 1) + 1)

Simplifying this equation
⇒ p(k + 1) : (1 + 1/1)(1 + 1/2)(1 + 1/3) ………….. (1 + 1/(k + 1)) = k + 2

Let’s simplify Left Hand Side of this equation
⇒ (1 + 1/1)(1 + 1/2)(1 + 1/3) ………….. (1 + 1/(k + 1))
⇒ (1 + 1/1)(1 + 1/2)(1 + 1/3) ………….. (1 + 1/k)(1 + 1/(k + 1))

Using [Equation 1] to replace value in above equation

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



15. 12 + 32 + 52 + …….. + (2n – 1)2 = n(2n – 1)(2n + 1)/3

Let’s suppose that P(n) : 12 + 32 + 52 + …….. + (2n – 1)2 = n(2n – 1)(2n + 1)/3

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = (2 × 1 – 1)2 = (2 – 1)2 = 12 = 1
Right Hand Side of P(n) = 1(2 × 1 – 1)(2 × 1 + 1)/3 = 1(2 – 1)(2 + 1)/3 = 3/3 = 1

Left Hand Side of P(n) = Right Hand Side of P(n) = 1

Thus P(m) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 12 + 32 + 52 + …….. + (2k – 1)2 = k(2k – 1)(2k + 1)/3 [Equation 1]

We need to prove that P(k + 1) is also True
⇒ p(k + 1) : 12 + 32 + 52 + …….. + (2(k + 1) – 1)2 = (k + 1)(2(k + 1) – 1)(2(k + 1) + 1)/3

Simplifying this equation
⇒ p(k + 1) : 12 + 32 + 52 + …….. + (2k + 2 – 1)2 = (k + 1)(2k + 2 – 1)(2k + 2 + 1)/3

⇒ p(k + 1) : 12 + 32 + 52 + …….. + (2k + 1)2 = (k + 1)(2k + 1)(2k + 3)/3
We need to prove that this equation is True

Let’s start with simplifying Left Hand Side of this equation
⇒ 12 + 32 + 52 + …….. + (2k + 1)2

This can be rewritten as
⇒ 12 + 32 + 52 + …….. + (2k – 1)2 + (2k + 1)2

Replacing value using [Equation 1]

\begin{equation} \frac{k(2k – 1)(2k + 1)}{3} + (2k + 1)^{2} \\ (2k + 1) \left[ \frac{k(2k – 1)}{3} + (2k + 1) \right] \\ (2k + 1) \left[ \frac{k(2k – 1) + 3(2k + 1)}{3} \right] \\ (2k + 1) \left[ \frac{2k^{2} – k + 6k + 3}{3} \right] \\ (2k + 1) \left[ \frac{2k^{2} + 5k + 3}{3} \right] \\ (2k + 1) \left[ \frac{2k^{2} + 2k + 3k + 3}{3} \right] \\ (2k + 1) \left[ \frac{2k(k + 1) + 3(k + 1)}{3} \right] \\ \frac{(2k + 1)(2k + 3)(k + 1)}{3} \\ \text{Which is equals to Right Hand Side of P(k + 1)} \end{equation}

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


16. 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3n – 2)(3n + 1) = n/(3n + 1)

Let’s suppose that P(n) : 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3n – 2)(3n + 1) = n/(3n + 1)

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1/(3 × 1 – 2)(3 × 1 + 1) = 1/(3 – 2)(3 + 1) = 1/(1)(4) = 1/4

Right Hand Side of P(n) = 1/(3 × 1 + 1) = 1/(3 + 1) = 1/4

Left Hand Side of P(n) = Right Hand Side of P(n) = 1/4

Thus P(m) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Suppose that P(k) is True
⇒ P(k) : 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3k – 2)(3k + 1) = k/(3k + 1) [Equation 1]

We need to prove that P(k + 1) is also True
⇒ P(k + 1) : 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3(k + 1) – 2)(3(k + 1) + 1) = (k + 1)/(3(k + 1) + 1)

Simplifying this equation
⇒ P(k + 1) : 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3k + 3 – 2)(3k + 3 + 1) = (k + 1)/(3k + 3 + 1)

⇒ P(k + 1) : 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3k + 1)(3k + 4) = (k + 1)/(3k + 4)
We need to prove that this equation is True

Let’s simplify Left Hand Side of this equation
⇒ 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3k + 1)(3k + 4)

⇒ 1/1.4 + 1/4.7 + 1/7.10 + ………. + 1/(3k – 2)(3k + 1) + 1/(3k + 1)(3k + 4)

Replacing value in this equation using [Equation 1]
⇒ k/(3k + 1) + 1/(3k + 1)(3k + 4)

\begin{equation} \frac{k}{(3k + 1)} + \frac{1}{(3k + 1)(3k + 4)} \\ \frac{1}{(3k + 1)} \left[ k + \frac{1}{(3k + 4)} \right] \\ \frac{1}{(3k + 1)} \left[ \frac{k(3k + 4) + 1}{(3k + 4)} \right] \\ \frac{1}{(3k + 1)} \left[ \frac{3k^{2} + 4k + 1}{(3k + 4)} \right] \\ \frac{1}{(3k + 1)} \left[ \frac{3k^{2} + 3k + k + 1}{(3k + 4)} \right] \\ \frac{1}{(3k + 1)} \left[ \frac{3k(k + 1) + 1(k + 1)}{(3k + 4)} \right] \\ \frac{1}{(3k + 1)} \left[ \frac{(3k + 1)(k + 1)}{(3k + 4)} \right] \\ \frac{k + 1}{(3k + 4)} \\ \text{which is equals to Right Hand Side of P(k + 1)} \end{equation}

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


17. 1/3.5 + 1/5.7 + 1/7.9 + ……….. + 1/(2n + 1)(2n + 3) = n/3(2n + 3)

Let’s suppose that P(n) : 1/3.5 + 1/5.7 + 1/7.9 + ……….. + 1/(2n + 1)(2n + 3) = n/3(2n + 3)

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1/(2 × 1 + 1)(2 × 1 + 3) = 1/(2 + 1)(2 + 3) = 1/(3)(5) = 1/15

Right Hand Side of P(n) = 1/3(2 × 1 + 3) = 1/3(2 + 3) = 1/3(5) = 1/55

Left Hand Side of P(n) = Right Hand Side of P(n) = 1/15

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : 1/3.5 + 1/5.7 + 1/7.9 + ……….. + 1/(2k + 1)(2k + 3) = k/3(2k + 3) [Equation 1]

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : 1/3.5 + 1/5.7 + 1/7.9 + ……….. + 1/(2(k + 1) + 1)(2(k + 1) + 3) = (k + 1)/3(2(k + 1) + 3)

Simplifying this
⇒ P(k + 1) : 1/3.5 + 1/5.7 + 1/7.9 + ……….. + 1/(2k + 3)(2k + 5) = (k + 1)/3(2k + 5)
We need to prove that this equation is True

Let’s start with simplifying Left Hand Side of this equation
⇒ 1/3.5 + 1/5.7 + 1/7.9 + ……….. + 1/(2k + 3)(2k + 5)

⇒ 1/3.5 + 1/5.7 + 1/7.9 + ……….. + 1/(2k + 1)(2k + 3) + 1/(2k + 3)(2k + 5)

Replacing value in this equation using [Equation 1]

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


18. 1 + 2 + 3 + ……… + n < 1/8(2n + 1)2

Let’s suppose that P(n) : 1 + 2 + 3 + ……… + n < 1/8(2n + 1)2

Proving P(n) for n = 1

For n = 1

Left Hand Side of P(n) = 1
Right Hand Side of P(n) = 1/8(2 × 1 + 1)2 = 1/8(2 + 1)2 = 1/8(3)2 = 1/8(9) = 1/72
Right Hand Side of P(n) = 1/72

Left Hand Side of P(n) = 1 < Right Hand Side of P(n) = 1/72

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : 1 + 2 + 3 + ……… + k < 1/8(2k + 1)2 [Equation 1]

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : 1 + 2 + 3 + ……… + (k + 1) < 1/8(2(k + 1) + 1)2

⇒ P(k + 1) : 1 + 2 + 3 + ……… + (k + 1) < 1/8(2k + 3)2
we need to prove that this equation is True

Using [Equation 1]
1 + 2 + 3 + ……… + k < 1/8(2k + 1)2

Adding (k + 1) to both sides of this equation

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



19. n(n + 1)(n + 5) is a multiple of 3

Let’s suppose that P(n) : n(n + 1)(n + 5) is a multiple of 3

Proving P(n) for n = 1

For n = 1

P(1) : 1(1 + 1)(1 + 5) is a multiple of 3
P(1) : 2 × 6 is a multiple of 3
P(1) : 12 is a multiple of 3 is True

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : k(k + 1)(k + 5) is a multiple of 3
⇒ P(k) : k(k2 + 5k + k + 5) is a multiple of 3
⇒ P(k) : k(k2 + 6k + 5) is a multiple of 3

Let’s suppose that m is a natural number
⇒ k(k2 + 6k + 5) = 3m
k3 + 6k2 + 5k = 3m
k3 = – 6k2 – 5k + 3m

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : (k + 1)((k +1) + 1)(k +1) + 5) is a multiple of 3

⇒ P(k + 1) : (k + 1)(k + 2)(k + 6) is a multiple of 3

⇒ P(k + 1) : (k + 1)(k2 + 6k + 2k + 12) is a multiple of 3

⇒ P(k + 1) : (k + 1)(k2 + 8k + 12) is a multiple of 3

⇒ P(k + 1) : k(k2 + 8k + 12) + 1(k2 + 8k + 12) is a multiple of 3

⇒ P(k + 1) : k3 + 8k2 + 12k + k2 + 8k + 12 is a multiple of 3

⇒ P(k + 1) : k3 + 9k2 + 20k + 12 is a multiple of 3
We need to prove that this equation is True

Replacing k3 = – 6k2 – 5k + 3m
⇒ P(k + 1) : – 6k2 – 5k + 3m + 9k2 + 20k + 12 = 3k2 + 15k + 3m + 12

⇒ P(k + 1) : 3k2 + 15k + 3m + 12

⇒ P(k + 1) : 3(k2 + 5k + m + 4)
Which is a multiple of 3

Thus we started with assuming that P(k) is True and we proved that P(k + 1) is also True

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


20. 102n – 1 + 1 is divisible by 11

Let’s suppose that P(n) : 102n – 1 + 1 is divisible by 11

Proving P(n) for n = 1

For n = 1

P(1) : 102×1 – 1 + 1 = 102 – 1 + 1 = 10 + 1 = 11
P(1) : 11 is divisible by 11

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : 102k – 1 + 1 is divisible by 11

Let’s suppose that m is a natural number
Thus above equation can be rewritten as
102k – 1 + 1 = 11m
102k – 1 = 11m – 1

102k/10 = 11m – 1

102k = 10(11m – 1) [Equation 1]

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : 102(k + 1) – 1 + 1 is divisible by 11

⇒ P(k + 1) : 102k + 2 – 1 + 1 is divisible by 11

⇒ P(k + 1) : 102k + 1 + 1 is divisible by 11
We need to prove that this equation is True

Let’s simplify P(k + 1)
⇒ 102k + 1 + 1

102k × 10 + 1

Replacing value of 102k using [Equation 1]
10(11m -1) × 10 + 1

100(11m – 1) + 1

1100m – 100 + 1

1100m – 99

11(100m – 9) is clearly divisible by 11

Thus if P(k) is True then P(k + 1) is also True

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



21. x2n – y2n is divisible by x + y

Let’s suppose that P(n) : x2n – y2n is divisible by x + y

Proving P(n) for n = 1

For n = 1

P(1) : x2 × 1 – y2 × 1 = x2 – y2 = (x – y)(x + y)

P(1) : (x – y)(x + y) is clearly divisible by (x + y)

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : x2k – y2k is divisible by x + y

Let’s suppose that m is a Natural Number
⇒ x2k – y2k = m(x + y)

x2k – y2k = mx + my

x2k = mx + my + y2k

x2k = m(x + y) + y2k [Equation 1]

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : x2(k + 1) – y2(k + 1) is divisible by x + y

P(k + 1) : x2(k + 1) – y2(k + 1) = x2k + 2 – y2k + 2

P(k + 1) : x2k + 2 – y2k + 2 = x2k x2 – y2k + 2

P(k + 1) : x2k x2 – y2k + 2

Replacing x2k value in [Equation 1]
P(k + 1) : (m(x + y) + y2k) x2 – y2k + 2

P(k + 1) : mx2(x + y) + y2k x2 – y2k + 2

P(k + 1) : mx2(x + y) + y2k(x2 – y2)

P(k + 1) : mx2(x + y) + y2k(x – y)(x + y)

P(k + 1) : (x + y)[mx2 + y2k(x – y)]
Which is clearly divisible by (x + y)

Thus if P(k) is True then P(k + 1) is also True

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


22. 32n + 2 – 8n – 9 is divisible by 8

Let’s suppose that P(n) : 32n + 2 – 8n – 9 is divisible by 8

Proving P(n) for n = 1

For n = 1

P(1) : 32(1) + 2 – 8(1) – 9 = 32 + 2 – 8 – 9 = 34 – 17 = 81 – 17 = 64

P(1) : 64 which is divisible by 8

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : 32k + 2 – 8k – 9 is divisible by 8

Let’s suppose that m is a Natural Number
32k + 2 – 8k – 9 = 8m

32k 32 – 8k – 9 = 8m

32k 32 = 8k + 9 + 8m

32k 9 = 8k + 9 + 8m

32k = (8k + 9 + 8m)/9 [Equation 1]

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : 32(k + 1) + 2 – 8(k + 1) – 9 is divisible by 8

Simplifying this
P(k + 1) : 32k + 2 + 2 – 8k – 8 – 9 = 32k + 4 – 8k – 17

P(k + 1) : 32k + 4 – 8k – 17

P(k + 1) : 32k 34 – 8k – 17

Replacing 32k value in this equation using [Equation 1]
P(k + 1) : (8k + 9 + 8m)/9 × 34 – 8k – 17

P(k + 1) : (8k + 9 + 8m)/9 × 81 – 8k – 17

P(k + 1) : 9(8k + 9 + 8m) – 8k – 17

P(k + 1) : 72k + 81 + 72m – 8k – 17 = 80k + 72m + 64

P(k + 1) : 80k + 72m + 64

P(k + 1) : 8(10k + 9m + 8)
Which is clearly divisible by 8

Hence if P(k) is True then P(k + 1) is also True

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.



23. 41n – 14n is a multiple of 27

Let’s suppose that P(n) : 41n – 14n is a multiple of 27

Proving P(n) for n = 1

For n = 1

P(1) : 411 – 141 = 41 – 14 = 27

P(1) : 27 is a multiple of 27

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : 41k – 14k is a multiple of 27

Let’s suppose that m is a Natural Number
⇒ 41k – 14k = 27m

41k = 27m + 14k [Equation 1]

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : 41k + 1 – 14k + 1 is a multiple of 27
We need to prove that this equation is True

⇒ 41k + 1 – 14k + 1

⇒ 41k 41 – 14k + 1

Replacing value of 41k in this equation using [Equation 1]
⇒ (27m + 14k) 41 – 14k + 1

⇒ 27m × 41 + 14k × 41 – 14k + 1

⇒ 27m × 41 + 14k × 41 – 14k × 14

⇒ 27m × 41 + 14k (41 – 14)

⇒ 27m × 41 + 14k × 27

⇒ 27(41m + 14k)
Which is clearly divisible by 27

Hence if P(k) is True then P(k + 1) is also True

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.


24. (2n + 7) < (n + 3)2

Let’s suppose that P(n) : (2n + 7) < (n + 3)2

Proving P(n) for n = 1

For n = 1

P(1) : (2 × 1 + 7) < (1 + 3)2

P(1) : (2 + 7) < 42

P(1) : 9 < 16

Thus P(n) is True for n = 1

Proving P(k + 1) is True if P(k) is True

Let’s suppose that P(k) is True
⇒ P(k) : (2k + 1) < (k + 3)2 [Equation 1]

Now we need to prove that P(k + 1) is also True
⇒ P(k + 1) : (2(k + 1) + 1) < ((k + 1) + 3)2

⇒ P(k + 1) : (2k + 2 + 1) < (k + 1 + 3)2

⇒ P(k + 1) : (2k + 3) < (k + 4)2
We need to prove that this in-equality is True

Using [Equation 1]
(2k + 1) < (k + 3)2

Adding 2 to both side of this in-equality
(2k + 1) + 2 < (k + 3)2 + 2

2k + 3 < (k + 3)2 + 2

2k + 3 < k2 + 2(k)(3) + 32 + 2

2k + 3 < k2 + 6k + 9 + 4

2k + 3 < k2 + 6k + 13
We know that k2 + 6k + 13 is less than k2 + 8k + 16

⇒ 2k + 3 < k2 + 6k + 13 < k2 + 8k + 16

⇒ 2k + 3 < k2 + 8k + 16

⇒ 2k + 3 < (k + 4)2
Which is same as P(k + 1)

Hence if P(k) is True then P(k + 1) is also True

⇒ If P(k) is True then P(k + 1) is also True

⇒ By Principle of Mathematical Induction, P(n) is True for any value of n such that n ∈ N.






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