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Question Number 65587 by naka3546 last updated on 31/Jul/19

$$Provethat$$$$1^3+2^3+3^3+\dots +n^3={(1+2+3+\dots +n)}^2$$

Commented by naka3546 last updated on 31/Jul/19

$$NousingMathematicalInduction.$$

Commented by Tanmay chaudhury last updated on 31/Jul/19

Commented by Tanmay chaudhury last updated on 31/Jul/19

$$S_3=3\int S_{2}dn+n\times B_3$$$$=3\int \frac{n(n+1)(2n+1)}{6}dn+n\times 0$$$$=\frac{1}{2}\int n(2n^2+3n+1)dn$$$$=\frac{1}{2}\int (2n^3+3n^2+n)dn$$$$\frac{1}{2}(\frac{2n^4}{4}+\frac{3n^3}{3}+\frac{n^2}{2})$$$$\frac{1}{2}\left(\frac{6n^4+12n^3+6n^2}{12}\right)$$$$\frac{1}{4}(n^4+2n^3+n^2)$$$$\frac{n^2}{4}(n^2+2n+1)$$$${\left\{\frac{n(n+1)}{2}\right\}}^{2}Answer$$

Commented by Tanmay chaudhury last updated on 31/Jul/19

Answered by som(math1967) last updated on 31/Jul/19

$$lets=1^3+2^3+3^3+\dots .+n^3$$$$now{n}^4-{(n-1)}^4=4n^3-6n^2+4n-1$$$$\therefore 1^4-0=4.1^3-6.1^2+4.1-1$$$$2^4-1^4=4.2^3-6.2^2+4.2-1$$$$\dots \dots \dots$$$$n^4-{(n-1)}^4=4n^3-6n^2+4n-1$$$$add$$$$n^4=4.(1^3+2^3+\dots ..n^3)-6(1^2+2^2+\dots n^2)$$$$+4(1+2+..n)-1\times n$$$$n^4=4s-6\times \frac{n(n+1)(2n+1)}{6}+4\times \frac{n(n+1)}{2}-n$$$$4s=n^4+n(n+1)(2n+1)-2n(n+1)+n$$$$4s=n(n^3+1)+n(n+1)(2n+1-2)$$$$4s=n(n+1)(n^2-n+1)+n(n+1)(2n-1)$$$$4s=n(n+1)(n^2-n+1+2n-1)$$$$4s=n(n+1)n(n+1)$$$$$$$$\therefore s={\left\{\frac{n(n+1)}{2}\right\}}^2$$$$\therefore s={(1+2+3+\dots \dots .n)}^2$$

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