Prove that 1^3 + 2^3 + 3^3 + … + n^3 = (1+2+3+...+n)^2


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

$P r o v e t h a t$ $1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = {(1 + 2 + 3 + . . . + n)}^{2}$
Commented by naka3546 last updated on 31/Jul/19

$N o u s i n g M a t h e m a t i c a l I n d u c t i o n .$
Commented by Tanmay chaudhury last updated on 31/Jul/19

Commented by Tanmay chaudhury last updated on 31/Jul/19
$S_3 =3∫S_2 dn+n×B_3 =3∫((n(n+1)(2n+1))/6)dn+n×0 =(1/2)∫n(2n^2 +3n+1)dn =(1/2)∫(2n^3 +3n^2 +n) dn (1/2)(((2n^4 )/4)+((3n^3 )/3)+(n^2 /2)) (1/2)(((6n^4 +12n^3 +6n^2 )/(12))) (1/4)(n^4 +2n^3 +n^2 ) (n^2 /4)(n^2 +2n+1) {((n(n+1))/2)}^2 Answer$
$S_{3} = 3 \int S_{2} d n + n \times B_{3}$ $= 3 \int \frac{n (n + 1) (2 n + 1)}{6} d n + n \times 0$ $= \frac{1}{2} \int n (2 n^{2} + 3 n + 1) d n$ $= \frac{1}{2} \int (2 n^{3} + 3 n^{2} + n) d n$ $\frac{1}{2} (\frac{2 n^{4}}{4} + \frac{3 n^{3}}{3} + \frac{n^{2}}{2})$ $\frac{1}{2} (\frac{6 n^{4} + 12 n^{3} + 6 n^{2}}{12})$ $\frac{1}{4} (n^{4} + 2 n^{3} + n^{2})$ $\frac{n^{2}}{4} (n^{2} + 2 n + 1)$ ${\frac{n (n + 1)}{2}}^{2} A n s w e r$
Commented by Tanmay chaudhury last updated on 31/Jul/19

	Answered by som(math1967) last updated on 31/Jul/19
	$let s=1^3 +2^3 +3^3 +....+n^3 now n^4 −(n−1)^4 =4n^3 −6n^2 +4n−1 ∴1^4 −0=4.1^3 −6.1^2 +4.1 −1 2^4 −1^4 =4.2^3 −6.2^2 +4.2−1 ......... n^4 −(n−1)^4 =4n^3 −6n^2 +4n−1 add n^4 =4.(1^3 +2^3 +.....n^3 )−6(1^2 +2^2 +...n^2 ) +4(1+2+..n)−1×n n^4 =4s−6×((n(n+1)(2n+1))/6) +4×((n(n+1))/2)−n 4s=n^4 +n(n+1)(2n+1) −2n(n+1)+n 4s=n(n^3 +1) +n(n+1)(2n+1−2) 4s=n(n+1)(n^2 −n+1)+n(n+1)(2n−1) 4s=n(n+1)(n^2 −n+1+2n−1) 4s=n(n+1)n(n+1) ∴s={((n(n+1))/2)}^2 ∴s=(1+2+3+.......n)^2$
	$l e t s = 1^{3} + 2^{3} + 3^{3} + . . . . + n^{3}$ $n o w n^{4} - {(n - 1)}^{4} = 4 n^{3} - 6 n^{2} + 4 n - 1$ $∴ 1^{4} - 0 = 4 . 1^{3} - 6 . 1^{2} + 4.1 - 1$ $2^{4} - 1^{4} = 4 . 2^{3} - 6 . 2^{2} + 4.2 - 1$ $. . . . . . . . .$ $n^{4} - {(n - 1)}^{4} = 4 n^{3} - 6 n^{2} + 4 n - 1$ $a d d$ $n^{4} = 4 . (1^{3} + 2^{3} + . . . . . n^{3}) - 6 (1^{2} + 2^{2} + . . . n^{2})$ $+ 4 (1 + 2 + . . n) - 1 \times n$ $n^{4} = 4 s - 6 \times \frac{n (n + 1) (2 n + 1)}{6} + 4 \times \frac{n (n + 1)}{2} - n$ $4 s = n^{4} + n (n + 1) (2 n + 1) - 2 n (n + 1) + n$ $4 s = n (n^{3} + 1) + n (n + 1) (2 n + 1 - 2)$ $4 s = n (n + 1) (n^{2} - n + 1) + n (n + 1) (2 n - 1)$ $4 s = n (n + 1) (n^{2} - n + 1 + 2 n - 1)$ $4 s = n (n + 1) n (n + 1)$ $∴ s = {\frac{n (n + 1)}{2}}^{2}$ $∴ s = {(1 + 2 + 3 + . . . . . . . n)}^{2}$
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