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Prove by method of induction, for all n ∈ N: 12 + 22 + 32 + .... + n2 = n(n+1)(2n+1)6

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प्रश्न

Prove by method of induction, for all n ∈ N:

12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`

बेरीज
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उत्तर

Let P(n) ≡ 12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`, for all n ∈ N

Step I:

Put n = 1

L.H.S. = 12 = 1

R.H.S. =  `(1(1 + 1)[2(1) + 1])/6 = 6/6` = 1

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1

Step II:

Let us consider that P(n) is true for n = k

∴ 12 + 22 + 32 + .... + k2 = `("k"("k" + 1)(2"k" + 1))/6`           ...(i)

Step III:

We have to prove that P(n) is true for n = k + 1

i.e., to prove that

12 + 22 + 32 + …. + (k + 1)2

= `(("k" + 1)[("k" + 1) + 1][2("k" + 1) + 1])/6`

= `(("k" + 1)("k" + 2)(2"k" + 3))/6`

L.H.S. = 12 + 22 + 32 + …. + (k + 1)

= 12 + 22 + 32 + …. + k2 + (k + 1)2

= `("k"("k" + 1)(2"k" + 1))/6 + ("k" + 1)^2`             ...[From (i)]

= `("k" + 1)[("k"(2"k" + 1))/6 + ("k" + 1)]`

= `("k" + 1)[(2"k"^2 + "k" + 6"k" + 6)/6]`

= `(("k" + 1)(2"k"^2 + 7"k" + 6))/6`

= `(("k" + 1)("k" + 2)(2"k" + 3))/6`

= R.H.S.

∴ P(n) is true for n = k + 1

Step IV:

From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6` for all n ∈ N.

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पाठ 4: Methods of Induction and Binomial Theorem - Exercise 4.1 [पृष्ठ ७३]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 3 | पृष्ठ ७३

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