Question

Using the principle of mathematical induction, prove each of the following for all n ϵ N:

3.2² + 3² .2³ + 3³ .2⁴ + …. + 3ⁿ .2ⁿ⁺¹ = \(\frac{12}5\) (6ⁿ – 1).

Answer 1

To Prove:

3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. + 3ⁿ  x 2ⁿ⁺¹ = \(\frac{12}5\) (6ⁿ – 1)

Let us prove this question by principle of mathematical induction (PMI)

Let P(n): 

3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. + 3ⁿ  x 2ⁿ⁺¹

For n = 1 

LHS = 3 x 2² = 12 

RHS = \((\frac{12}5)\) x (61 - 1)

= \(\frac{12}5\) x 5 = 12 

Hence, LHS = RHS 

P(n) is true for n = 1 

Assume P(k) is true

3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. + 3^k  x 2^k+1 = \(\frac{12}5\) (6^k – 1)....(1)

We will prove that P(k + 1) is true

We have to prove P(k + 1) from P(k) ie (2) from (1) 

From (1)

3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. x + 3^k  x 2^k+1 = \(\frac{12}5\) (6^k – 1)

Adding 3^k+1 x 2^k+2  both sides

3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. + 3^k  x 2^k+1 + 3^{k + 1} x 2^{k + 2} = \(\frac{12}5\)(6^k - 1) + 3^{k + 1} x 2^{k + 2}

3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. + 3^k  x 2^k+1 + 3^{k + 1} x 2^{k + 2} = \(\frac{12}5\)(6^k +1) - \(\frac{12}5\)

Which is the same as P(k + 1) 

Therefore, P (k + 1) is true whenever P(k) is true. 

By the principle of mathematical induction, P(n) is true for × 

Where n is a natural number 

Put k = n - 1

3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. + 3ⁿ  x 2^{n+1 =} \(\frac{12}5\)(6ⁿ) - \(\frac{12}5\)

 3 x 2² + 3² x 2³ + 3³ x 2⁴ + …. + 3ⁿ  x 2^{n+1 =} \(\frac{12}5\)(6ⁿ - 1) 

Hence proved