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By the principle of mathematical induction, prove the following: 13 + 23 + 33 + ….. + n3 = nn + 1n2(n + 1)24 for all x ∈ N. - Business Mathematics and Statistics

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प्रश्न

By the principle of mathematical induction, prove the following:

13 + 23 + 33 + ….. + n3 = `("n"^2("n + 1")^2)/4` for all x ∈ N.

बेरीज

उत्तर

Let P(n) be the statement 13 + 23 + 33 + ….. + n3 = `("n"^2("n + 1")^2)/4` for all x ∈ N.

i.e., p(n) = 13 + 23 + …… + n3 = `("n"^2("n + 1")^2)/4` for all x ∈ N

Put n = 1

LHS = 13 = 1

RHS = `(1^2(1 + 1)^2)/4`

`= (1 xx 2^2)/4`

`= 4/4` = 1

∴ P(1) is true.

Assume that P(n) is true n = k

P(k): 13 + 23 + …… + k3 = `(k^2(k + 1)^2)/4`

To prove P(k + 1) is true.

i.e., to prove 13 + 23 + ……. + k3 + (k + 1)3 = `((k + 1)^2 ((k+1)+1)^2)/4 = ((k + 1)^2(k + 2)^2)/4`

Consider 13 + 23 + …… + k3 + (k + 1)3 = `(k^2(k + 1)^2)/4 + (k + 1)^3`

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

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

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

⇒ P(k + 1) is true, whenever P(k) is true.

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

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Mathematical Induction
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Algebra - Exercise 2.5 [पृष्ठ ४१]

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