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Question
Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by
Sn = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`
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Solution
Here P(n): Sn = `{{:((n(n + 1)^2)/2",", "when n is even"),((n^2(n + 1))/2",", "when n is odd"):}`
Also, note that any term Tn of the series is given by
Tn = `{{:(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) : Sk + 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.
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