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Question
By the principle of mathematical induction, prove that, for n ≥ 1
13 + 23 + 33 + ... + n3 = `(("n"("n" + 1))/2)^2`
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Solution
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 + ... + k3 + (k + 1)3
= `[("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.
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