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By the principle of mathematical induction, prove that, for n ≥ 1 13 + 23 + 33 + ... + n3 = nn(n(n+1)2)2 - Mathematics

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प्रश्न

By the principle of mathematical induction, prove that, for n ≥ 1
13 + 23 + 33 + ... + n3 = `(("n"("n" + 1))/2)^2`

योग
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उत्तर

P(n) = 13 + 23 + 33 + ... + n3 = `(("n"("n" + 1))/2)^2`

For n = 1

P(1) = 1

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

⇒ 1 = 1

∴ P(1) is true

Let P(n) be true for n = k

∴ P(k) = 13 + 23 + 33 + ... + k3 

= `[("k"("k" + 1))/2]^2`    ......(i)

From n = k + 1

P(k + 1) = 13 + 23 + 33 + ... + k+ (k + 1)

= `[("k"("k" + 1))/2]^2 + ("k" + 1)^3`   ......[Using (i)]

= `("k" + 1)^2 ["k"^2/4 + "k" + 1]` 

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

=`(("k" + 1)^2("k" + 2)^2)/4`

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

∴ P(k+ 1) is true.

Thus P(k) is true

⇒ (k + 1) is true.

Hence by principle of mathematical induction

P(n) 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 1 | पृष्ठ १९६

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