Question

Ma poate ajuta cineva cu demonstratia:

1²+3²+5²+...+(2n-1)²=[n(4n²-1)]/3

Dar nu prin inductie, ci folosind sume remarcabile

Answer 1

$1^2+3^2+5^2+...+(2n-1)^2 =$
[tex]\\=\sum\limits_{k=1}^n (2k-1)^2 =\sum\limits_{k=1}^n (4k^2-4k+1) = \sum\limits_{k=1}^n 4k^2-\sum\limits_{k=1}^n 4k+\sum\limits_{k=1}^n 1 = \\ \\ = 4\cdot\sum\limits_{k=1}^n k^2-4\cdot\sum\limits_{k=1}^n k+\sum\limits_{k=1}^n 1 = \\ \\ = 4\cdot (1^2+2^2+3^2+...+n^2)-4\cdot (1+2+3+...+n)+n = \\ \\ = 4\cdot\dfrac{n(n+1)(2n+1)}{6}-4\cdot \dfrac{n(n+1)}{2}+n = \\ \\ [/tex]

$= 2\cdot \dfrac{n(n+1) (2n+1)}{3}-2 n (n+1)+n = \\ \\ = \dfrac{2 n (n+1) (2n+1)-3\cdot 2 n\ (n+1)+3\cdot n}{3} = \\ \\ =\dfrac{n\cdot\Big[2(n+1)(2n+1)-6(n+1)+3\Big]}{3} = \\ \\ = \dfrac{n\cdot \Big[2\cdot(2n^2+n+2n+1)-6n-6+3\Big] }{3} = \\ \\ = \dfrac{n\cdot \Big(4n^2+6n+2-6n-3\Big)}{3} = \\ \\ = \dfrac{n(4n^2-1)}{3}$