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

**1. 1 + 3 + 3 ^{2} + ……….. + 3^{n-1} = (3^{n} – 1)/2**

Let’s suppose that **P(n) : 1 + 3 + 3 ^{2} + ……….. + 3^{n-1} = (3^{n} – 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 + 3^{2} + ……….. + 3^{k-1} = (3^{k} – 1)/2 is True [Equation 1]

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

⇒ P(k + 1) : 1 + 3 + 3^{2} + ……….. + 3^{(k + 1) – 1} = (3^{k + 1} – 1)/2

Simplifying this

⇒ P(k + 1) : 1 + 3 + 3^{2} + ……….. + 3^{k} = (3^{k + 1} – 1)/2 *(We need to prove this)* [Equation 2]

Using [Equation 1]

⇒ P(k) : 1 + 3 + 3^{2} + ……….. + 3^{k-1} = (3^{k} – 1)/2

Adding 3^{k} to both sides of above equation

So above equation can be written as

1 + 3 + 3^{2} + ……….. + 3^{k-1} + 3^{k} = (3^{k} – 1)/2 + 3^{k}

Let’s now simplify Right Hand Side of above equation

1 + 3 + 3^{2} + ……….. + 3^{k-1} + 3^{k} = (3^{k} – 1 + 2 × 3^{k})/2 = (3 × 3^{k} – 1)/2 = (3^{k + 1} – 1)/2

⇒ 1 + 3 + 3^{2} + ……….. + 3^{k-1} + 3^{k} = (3^{k + 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. 1 ^{3} + 2^{3} + 3^{3} + ………… + n^{3} = (n(n + 1)/2)^{2}**

Let’s suppose that

**P(n) : 1**

^{3}+ 2^{3}+ 3^{3}+ ………… + n^{3}= (n(n + 1)/2)^{2}*Proving P(n) for n = 1*

For n = 1

Left Hand Side of P(n) = 1^{3} = 1

Right Hand Side of P(n) = (1(1 + 1)/2)^{2} = (2/2)^{2} = 1^{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} + 2^{3} + 3^{3} + ………… + k^{3} = (k(k + 1)/2)^{2} is True [Equation 1]

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

Using [Equation 1]

⇒ P(k) : 1^{3} + 2^{3} + 3^{3} + ………… + k^{3} = (k(k + 1)/2)^{2}

Adding (k + 1)^{3} to both sides of this equation

⇒ P(k) : 1^{3} + 2^{3} + 3^{3} + ………… + k^{3} + (k + 1)^{3} = (k(k + 1)/2)^{2} + (k + 1)^{3}

Let’s simplify this

⇒ 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 + 2.3 ^{2} + 3.3^{3} + ………….. + n.3^{n} = [(2n – 1)3^{n+1} + 3]/4**

Let’s suppose that

**P(n) : 1.3 + 2.3**

^{2}+ 3.3^{3}+ ………….. + n.3^{n}= [(2n – 1)3^{n+1}+ 3]/4*Proving P(n) for n = 1*

For n = 1

Left Hand Side of P(n) = 1.3^{1} = 3

Right Hand Side of P(n) = [(2 × 1 – 1)3^{1 + 1} + 3]/4 = [(2 – 1)3^{2} + 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 + 2.3^{2} + 3.3^{3} + ………….. + k.3^{k} = [(2k – 1)3^{k+1} + 3]/4 [Equation 1]

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

⇒ P(k + 1) : 1.3 + 2.3^{2} + 3.3^{3} + ………….. + (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 + 2.3^{2} + 3.3^{3} + ………….. + k.3^{k}** + (k + 1).3

^{(k + 1)}

Using [Equation 1] replacing the value

⇒ [(2k – 1)3

^{k+1}+ 3]/4 + (k + 1).3

^{(k + 1)}

Simplifying this

Which is what we need to prove.

Hence it’s proved that**P(k + 1) : 1.3 + 2.3 ^{2} + 3.3^{3} + ………….. + (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)

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(4n ^{2} + 6n – 1)/3**

Let’s suppose that

**P(n) : 1.3 + 3.5 + 5.7 + ……… + (2n – 1)(2n + 1) = n(4n**

^{2}+ 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 × 1^{2} + 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(4k^{2} + 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(k^{2} + 2k + 1) + 6k + 6 – 1)/3

⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (k + 1)(4k^{2} + 8k + 4 + 6k + 6 – 1)/3

⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (k + 1)(4k^{2} + 14k + 9)/3

⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (4k^{3} + 14k^{2} + 9k + 4k^{2} + 14k + 9)/3

⇒ P(k + 1) : 1.3 + 3.5 + 5.7 + ……… + (2k + 1)(2k + 3) = (4k^{3} + 18k^{2} + 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(4k

^{2}+ 6k – 1)/3 + (2k + 1)(2k + 3)

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 + 2.2 ^{2} + 3.2^{3} + ………. + n.2^{n} = (n – 1)2^{n + 1} + 2**

Let’s suppose that

**P(n) : 1.2 + 2.2**

^{2}+ 3.2^{3}+ ………. + n.2^{n}= (n – 1)2^{n + 1}+ 2*Proving P(n) for n = 1*

For n = 1

Left Hand Side of P(n) = 1.2^{1} = 2

Right Hand Side of P(n) = (1 – 1)2^{1 + 1} + 2 = (0)2^{2} + 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 + 2.2^{2} + 3.2^{3} + ………. + k.2^{k} = (k – 1)2^{k + 1} + 2 [Equation 1]

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

⇒ P(k + 1) : 1.2 + 2.2^{2} + 3.2^{3} + ………. + (k + 1).2^{k + 1} = ((k + 1) – 1)2^{(k + 1) + 1} + 2

Simplifying this equation

⇒ P(k + 1) : 1.2 + 2.2^{2} + 3.2^{3} + ………. + (k + 1).2^{k + 1} = (k + 1 – 1)2^{k + 1 + 1} + 2

⇒ P(k + 1) : 1.2 + 2.2^{2} + 3.2^{3} + ………. + (k + 1).2^{k + 1} = k.2^{k+2} + 2*We need to prove this*

Let’s rewrite Left Hand Side of this equation

⇒ 1.2 + 2.2^{2} + 3.2^{3} + ………. + k.2^{k} + (k + 1).2^{k + 1}

Using [Equation 1] to rewrite this equation

⇒ (k – 1)2^{k + 1} + 2 + (k + 1).2^{k + 1}

⇒ 2^{k + 1}(k – 1 + k + 1) + 2

⇒ 2^{k + 1}(2k) + 2

⇒ k.2^{k+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 + …………. + 1/2 ^{n} = 1 – 1/2^{n}**

Let’s suppose that

**P(n) : 1/2 + 1/4 + 1/8 + …………. + 1/2**

^{n}= 1 – 1/2^{n}*Proving P(n) for n = 1*

For n = 1

Left Hand Side of P(n) = 1/2^{1} = 1/2

Right Hand Side of P(n) = 1 – 1/2^{1} = 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 + …………. + 1/2^{k} = 1 – 1/2^{k} [Equation 1]

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

⇒ P(k + 1) : 1/2 + 1/4 + 1/8 + …………. + 1/2^{k + 1} = 1 – 1/2^{k + 1}

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

Using [Equation 1] to replace value in the above 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 + ar ^{2} + ………. + ar^{n-1} = a(r^{n} – 1)/r – 1**

Let’s suppose that

**P(n) : a + ar + ar**

^{2}+ ………. + ar^{n-1}= a(r^{n}– 1)/r – 1*Proving P(n) for n = 1*

For n = 1

Left Hand Side of P(n) = ar^{1-1} = ar^{0} = a(1) = a

Right Hand Side of P(n) = a(r^{1} – 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 + ar^{2} + ………. + ar^{k-1} = a(r^{k} – 1)/(r – 1) [Equation 1]

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

⇒ P(k + 1) : a + ar + ar^{2} + ………. + ar^{(k + 1)-1} = a(r^{(k + 1)} – 1)/(r – 1)

Let’s simplify this equation

⇒ P(k + 1) : a + ar + ar^{2} + ………. + ar^{k} = 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 + ar^{2} + ………. + ar^{k-1} + ar^{k}

Replacing value in this equation using [Equation 1]

⇒ a(r^{k} – 1)/(r – 1) + ar^{k}

⇒ 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)/n ^{2}) = (n + 1)^{2}**

*Proving P(n) for n = 1*

For n = 1

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

Right Hand Side of P(n) = (1 + 1)^{2} = 2^{2} = 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

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

**15. 1 ^{2} + 3^{2} + 5^{2} + …….. + (2n – 1)^{2} = n(2n – 1)(2n + 1)/3**

Let’s suppose that

**P(n) : 1**

^{2}+ 3^{2}+ 5^{2}+ …….. + (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} = 1^{2} = 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) : 1^{2} + 3^{2} + 5^{2} + …….. + (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) : 1^{2} + 3^{2} + 5^{2} + …….. + (2(k + 1) – 1)^{2} = (k + 1)(2(k + 1) – 1)(2(k + 1) + 1)/3

Simplifying this equation

⇒ p(k + 1) : 1^{2} + 3^{2} + 5^{2} + …….. + (2k + 2 – 1)^{2} = (k + 1)(2k + 2 – 1)(2k + 2 + 1)/3

⇒ p(k + 1) : 1^{2} + 3^{2} + 5^{2} + …….. + (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

⇒ 1^{2} + 3^{2} + 5^{2} + …….. + (2k + 1)^{2}

This can be rewritten as

⇒ 1^{2} + 3^{2} + 5^{2} + …….. + (2k – 1)^{2} + (2k + 1)^{2}

Replacing value using [Equation 1]

⇒ 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)

⇒ 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]

⇒ 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

⇒ 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(k^{2} + 5k + k + 5) is a multiple of 3

⇒ P(k) : k(k^{2} + 6k + 5) is a multiple of 3

Let’s suppose that m is a natural number

⇒ k(k^{2} + 6k + 5) = 3m

k^{3} + 6k^{2} + 5k = 3m*k ^{3} = – 6k^{2} – 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)(k

^{2}+ 6k + 2k + 12) is a multiple of 3

⇒ P(k + 1) : (k + 1)(k

^{2}+ 8k + 12) is a multiple of 3

⇒ P(k + 1) : k(k

^{2}+ 8k + 12) + 1(k

^{2}+ 8k + 12) is a multiple of 3

⇒ P(k + 1) : k

^{3}+ 8k

^{2}+ 12k + k

^{2}+ 8k + 12 is a multiple of 3

⇒ P(k + 1) : k

^{3}+ 9k

^{2}+ 20k + 12 is a multiple of 3

*We need to prove that this equation is True*Replacing

*k*^{3}= – 6k^{2}– 5k + 3m⇒ P(k + 1) : – 6k

^{2}– 5k + 3m + 9k

^{2}+ 20k + 12 = 3k

^{2}+ 15k + 3m + 12

⇒ P(k + 1) : 3k

^{2}+ 15k + 3m + 12

⇒ P(k + 1) : 3(k

^{2}+ 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. 10 ^{2n – 1} + 1 is divisible by 11**

Let’s suppose that

**P(n) : 10**

^{2n – 1}+ 1 is divisible by 11*Proving P(n) for n = 1*

For n = 1

P(1) : 10^{2×1 – 1} + 1 = 10^{2 – 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) : 10^{2k – 1} + 1 is divisible by 11

Let’s suppose that m is a natural number

Thus above equation can be rewritten as

10^{2k – 1} + 1 = 11m

10^{2k – 1} = 11m – 1

10^{2k}/10 = 11m – 1** 10^{2k} = 10(11m – 1)** [Equation 1]

Now we need to prove that P(k + 1) is also True

⇒ P(k + 1) : 10

^{2(k + 1) – 1}+ 1 is divisible by 11

⇒ P(k + 1) : 10

^{2k + 2 – 1}+ 1 is divisible by 11

⇒ P(k + 1) : 10

^{2k + 1}+ 1 is divisible by 11

*We need to prove that this equation is True*

Let’s simplify P(k + 1)

⇒ 10

^{2k + 1}+ 1

10

^{2k}× 10 + 1

Replacing value of 10

^{2k}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. x ^{2n} – y^{2n} is divisible by x + y**

Let’s suppose that

**P(n) : x**

^{2n}– y^{2n}is divisible by x + y*Proving P(n) for n = 1*

For n = 1

P(1) : x^{2 × 1} – y^{2 × 1} = x^{2} – y^{2} = (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) : x^{2k} – y^{2k} is divisible by x + y

Let’s suppose that m is a Natural Number

⇒ x^{2k} – y^{2k} = m(x + y)

x^{2k} – y^{2k} = mx + my

x^{2k} = mx + my + y^{2k}** x^{2k} = m(x + y) + y^{2k}** [Equation 1]

Now we need to prove that P(k + 1) is also True

⇒ P(k + 1) : x

^{2(k + 1)}– y

^{2(k + 1)}is divisible by x + y

P(k + 1) : x

^{2(k + 1)}– y

^{2(k + 1)}= x

^{2k + 2}– y

^{2k + 2}

P(k + 1) : x

^{2k + 2}– y

^{2k + 2}= x

^{2k}x

^{2}– y

^{2k + 2}

P(k + 1) : x

^{2k}x

^{2}– y

^{2k + 2}

Replacing x

^{2k}value in [Equation 1]

P(k + 1) : (m(x + y) + y

^{2k}) x

^{2}– y

^{2k + 2}

P(k + 1) : mx

^{2}(x + y) + y

^{2k}x

^{2}– y

^{2k + 2}

P(k + 1) : mx

^{2}(x + y) + y

^{2k}(x

^{2}– y

^{2})

P(k + 1) : mx

^{2}(x + y) + y

^{2k}(x – y)(x + y)

P(k + 1) : (x + y)[mx

^{2}+ y

^{2k}(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. 3 ^{2n + 2} – 8n – 9 is divisible by 8**

Let’s suppose that

**P(n) : 3**

^{2n + 2}– 8n – 9 is divisible by 8*Proving P(n) for n = 1*

For n = 1

P(1) : 3^{2(1) + 2} – 8(1) – 9 = 3^{2 + 2} – 8 – 9 = 3^{4} – 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) : 3^{2k + 2} – 8k – 9 is divisible by 8

Let’s suppose that m is a Natural Number

3^{2k + 2} – 8k – 9 = 8m

3^{2k} 3^{2} – 8k – 9 = 8m

3^{2k} 3^{2} = 8k + 9 + 8m

3^{2k} 9 = 8k + 9 + 8m** 3^{2k} = (8k + 9 + 8m)/9** [Equation 1]

Now we need to prove that P(k + 1) is also True

⇒ P(k + 1) : 3

^{2(k + 1) + 2}– 8(k + 1) – 9 is divisible by 8

Simplifying this

P(k + 1) : 3

^{2k + 2 + 2}– 8k – 8 – 9 = 3

^{2k + 4}– 8k – 17

P(k + 1) : 3

^{2k + 4}– 8k – 17

P(k + 1) : 3

^{2k}3

^{4}– 8k – 17

Replacing 3

^{2k}value in this equation using [Equation 1]

P(k + 1) : (8k + 9 + 8m)/9 × 3

^{4}– 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. 41 ^{n} – 14^{n} is a multiple of 27**

Let’s suppose that

**P(n) : 41**

^{n}– 14^{n}is a multiple of 27*Proving P(n) for n = 1*

For n = 1

P(1) : 41^{1} – 14^{1} = 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) : 41^{k} – 14^{k} is a multiple of 27

Let’s suppose that m is a Natural Number

⇒ 41^{k} – 14^{k} = 27m

41^{k} = 27m + 14^{k} [Equation 1]

Now we need to prove that P(k + 1) is also True

⇒ P(k + 1) : 41^{k + 1} – 14^{k + 1} is a multiple of 27**We need to prove that this equation is True**

⇒ 41^{k + 1} – 14^{k + 1}

⇒ 41^{k} 41 – 14^{k + 1}

Replacing value of 41^{k} in this equation using [Equation 1]

⇒ (27m + 14^{k}) 41 – 14^{k + 1}

⇒ 27m × 41 + 14^{k} × 41 – 14^{k + 1}

⇒ 27m × 41 + 14^{k} × 41 – 14^{k} × 14

⇒ 27m × 41 + 14^{k} (41 – 14)

⇒ 27m × 41 + 14^{k} × 27

⇒ 27(41m + 14^{k})

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) < 4^{2}

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 < k^{2} + 2(k)(3) + 3^{2} + 2

2k + 3 < k^{2} + 6k + 9 + 4

2k + 3 < k^{2} + 6k + 13

We know that **k ^{2} + 6k + 13** is less than

**k**

^{2}+ 8k + 16⇒ 2k + 3 < k

^{2}+ 6k + 13 < k

^{2}+ 8k + 16

⇒ 2k + 3 < k

^{2}+ 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.