Question

Show by the Principle of Mathematical Induction that the sum S_n of the n term of the series 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² ... is given by

S_n = `{{:((n(n + 1)^2)/2",",  "if n is even"),((n^2(n + 1))/2",",  "if n is odd"):}`

Answer 1

Here P(n): S_n = `{{:((n(n + 1)^2)/2",",  "when n is even"),((n^2(n + 1))/2",",  "when n is odd"):}`

Also, note that any term T_n of the series is given by

T_n = `{{:(n^2, "if n is odd"),(2n^2, "if n is even"):}`

We observe that P(1) is true.

Since P(1): S₁ = 1²

= 1

= `(1.2)/2`

= `(1^2.(1 + 1))/2`

Assume that P(k) is true for some natural number k, i.e.

Case 1: When k is odd, then k + 1 is even.

We have P(k + 1) : S_{k + 1} = 1² + 2 × 2² + ... + k² + 2 × (k + 1)²

= `(k^2(k + 1))/2 + 2 xx (k + 1)^2`

= `((k + 1))/2 [k^2 + 4(k + 1)]`  .....(As k is odd, 1² + 2 × 2² + ... + k² = `k^2 ((k + 1))/2`)

= `(k + 1)/2 [k^2 + 4k + 4]`

= `(k + 1)/2 (k + 2)^2`

= `(k + 1) ([(k + 1) + 1]^2)/2`

So P(k + 1) is true.

Whenever P(k) is true in the case when k is odd.

Case 2: When k is even, then k + 1 is odd.

Now, P(k + 1): 1² + 2 × 2² + ... + 2.k² + (k + 1)²

= `(k(k + 1)^2)/2 + (k + 1)^2`  ......(As k is even, 1² + 2 × 2² + ... + 2k² = `k(k + 1)^2/2`)

= `((k + 1)^2 (k + 2))/2`

= `((k + 1)^2 ((k + 1) + 1))/2`

Therefore, P(k + 1) is true.

Whenever P(k) is true for the case when k is even.

Thus, P(k + 1) is true.

Whenever, P(k) is true for any natural numbers k.

Hence, P(n) true for all natural numbers.