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By the principle of mathematical induction, prove that, for n ≥ 112 + 32 + 52 + ... + (2n − 1)2 = nnnn(2n-1)(2n+1)3 - Mathematics

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प्रश्न

By the principle of mathematical induction, prove that, for n ≥ 1
12 + 32 + 52 + ... + (2n − 1)2 = `("n"(2"n" - 1)(2"n" + 1))/3`

योग
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उत्तर

Let P(n) = 12 + 32 + 52 + ... + (2n − 1)2 = `("n"(2"n" - 1)(2"n" + 1))/3`

For n = 1

P(1) = (2 × 1 − 1)2

= `(1(2 xx 1 - 1)(2 xx 1 + 1))/3`

⇒ 1 = `(1 xx 1 xx 3)/3`

∴ P(1) is true

Let P(n) be true for n = k

∴ P(k) = 12 + 32 + 52 + ... + (2k − 1)2 = `("n"(2"k" - 1)(2"k" + 1))/3`  ......(i)

For n = k + 1

R.H.S = `(("k" + 1)(2"k" + 1)(2"k" + 3))/3`

L.H.S = `("k"(2"k" - 1)(2"k" + 1))/3  ("k" + 1)^2` .....(Using (i)]

= `(2"k" + 1) [("k"(2"k" - 1))/3 + (2"k" + 1)]`

= `(2"k" + 1) [(2"k"^2- "k" + 6"k" + 3)/3]`

=`((2"k" - 1)(2"k"^2 + 5"k" + 3))/3`

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

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

∴ P(k + 1) is true

Thus P(k) is true

⇒ P(k + 1) is true.

Hence by principle of mathematical induction,

P(k) is true for all n ∈ N.

shaalaa.com
Mathematical Induction
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Combinatorics and Mathematical Induction - Exercise 4.4 [पृष्ठ १९६]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 4 Combinatorics and Mathematical Induction
Exercise 4.4 | Q 2 | पृष्ठ १९६

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