Question

Sa dem: 1^3+2^3+...+n^3=[n*(n+1)/2]^2. Nu prin inductie!!

Answer 1

$\sum\limits_{k=1}^n\Big(k^4-(k-1)^4\Big) = \sum\limits_{k=1}^n k^4-\sum\limits_{k=1}^n(k-1)^4 = \\ \\ = 1^4+2^4+3^4+...+n^4-\Big(0^4+1^4+2^4+3^4+...+(n-1)^4\Big) = \\ = 1^4+2^4+3^4+...+(n-1)^4+n^4-\Big(1^4+2^4+3^4+...+(n-1)^4\Big) = \\ = n^4 \\ \\\Rightarrow \boxed{\sum\limits_{k=1}^n\Big(k^4-(k-1)^4\Big) = n^4} (1)\\ \\ \\ k^4-(k-1)^4 = k^4-(k-1)^2\cdot (k-1)^2 = \\ =k^4-\Big((k^2-2k+1)(k^2-2k+1)\Big) = \\ = k^4-(k^4-2k^3+k^2-2k^3+4k^2-2k+k^2-2k+1) = \\ = k^4 - (k^4-4k^3+6k^2-4k+1) = \\ =4k^3-6k^2+4k-1$

$\Rightarrow \boxed{k^4-(k-1)^4 = 4k^3-6k^2+4k-1} (2)$

$\boxed{\text{Din (1) si (2) } \Rightarrow }$

$\sum\limits_{k=1}^n(4k^3-6k^2+4k-1) = n^4 \\ \sum\limits_{k=1}^n(4k^3) - \sum\limits_{k=1}^n(6k^2)+\sum\limits_{k=1}^n(4k) -\sum\limits_{k=1}^n 1 = n^4 \\ 4\sum\limits_{k=1}^nk^3-6 \sum\limits_{k=1}^nk^2+4\sum\limits_{k=1}^nk-\sum\limits_{k=1}^n 1 = n^4 \\4\sum\limits_{k=1}^nk^3 -6\cdot \dfrac{n(n+1)(2n+1)}{6}+4\cdot \dfrac{n(n+1)}{2}-n = n^4$
$4\sum\limits_{k=1}^nk^3 =n^4+n(n+1)(2n+1)-2n(n+1)+n \\ 4\sum\limits_{k=1}^nk^3 =n(n^3+1)+n(n+1)(2n+1)-2n(n+1) \\ 4\sum\limits_{k=1}^nk^3 = n\Big(n^3+1+(n+1)(2n+1)-2(n+1)\Big) \\ 4\sum\limits_{k=1}^n k^3 = n\Big((n+1)(n^2-n+1)+(n+1)(2n+1-2)\Big) \\ 4\sum\limits_{k=1}^nk^3 = n(n+1)\Big(n^2-n+1+2n+1-2\Big) \\ 4\sum\limits_{k=1}^n k^3 = n(n+1)(n^2+n)\\ 4\sum\limits_{k=1}^n k^3 = n(n+1)n(n+1) \\ 4\sum\limits_{k=1}^n k^3 = \Big(n(n+1)\Big)^2$
$\sum\limits_{k=1}^n k^3 = \dfrac{\Big(n(n+1)\Big)^2}{4} \\ \sum\limits_{k=1}^n k^3 = \left(\dfrac{n(n+1)}{2}\right)^2 \\ \\ \\\Rightarrow \boxed{\boxed{1^3+2^3+3^3+...+n^3 = \left(\dfrac{n(n+1)}{2}\right)^2 }}$