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Prove the following by using the principle of mathematical induction for all $n \in N $: \[\] $1^3+2^3+3^3+...+n^3= \bigg( \large\frac{n(n+1)}{2} \bigg)^2$

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Let the given statement be $P(n)$, i.e.,
$P(n) : 1^3+2^3+3^3+...+n^3= \bigg( \large\frac{n(n+1)}{2} \bigg)^2$
For $n = 1$, we have
$P(1):1^3 =1 =\bigg( \large\frac{ 1(1+1)}{2} \bigg)^2$$ = \bigg(\large\frac{1.2}{2} \bigg)$$ = 1^2=1$ which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$1^3+2^3+3^3+...+k^3= \bigg(\large\frac{k(k+1)}{2} \bigg)^2$-------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$1^3+2^3+3^3+...k^3+(k+1)^3$
$= \bigg(\large\frac{k(k+1)}{2} \bigg)^2$$+(k+1)^3 \qquad $ [ Using (i) ]
$= \large\frac{k^2(k+1)}{4} ^2$$+(k+1)^3 $
$ = \large\frac{k^2+(k+1)^2+4 (k+1)^3}{4}$
$ = \large\frac{(k+1)^2 \{ k^2+4(k+1) \}}{4}$
$ = \large\frac{(k+1)^2 \{ k^2+4k+4) \}}{4}$
$ = \large\frac{(k+1)^2 (k+2)^2}{4}$
$ = \large\frac{(k+1)^2 (k+1+1)^2}{4}$
$=(1^3+2^3+3^3+...k^3)+(k+1)^3 = \bigg(\large\frac{(k+1)(k+1+1)}{2} \bigg)^2$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered Apr 29, 2014 by thanvigandhi_1
 

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