हिंदी

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 = ,if n is even,if n is odd{n(n+1)22, if n

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प्रश्न

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"):}`

योग
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उत्तर

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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अध्याय 4: Principle of Mathematical Induction - Solved Examples [पृष्ठ ६७]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 4 Principle of Mathematical Induction
Solved Examples | Q 10 | पृष्ठ ६७

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