Question

2^2+6^2+...+(4n+2)^2= ?

Answer 1

Răspuns:

Explicație pas cu pas:

vom folosi formula:

$1^{2} +2^{2} +3^{2} +....+n^{2} =\frac{n(n+1)(2n+1)}{6}$

$2^{2} +6^{2} +...+(4n+2)^{2} =2^{2} [1^{2} +3^{2}+...+(2n+1)^{2} ]=\\  =2^{2} [1^{2} +2^{2} +3^{2} +4^{2} +...+(2n)^{2} +(2n+1)^{2} -2^{2} -4^{2}-...-(2n)^{2} ]=\\= 2^{2} [\frac{(2n+1)(2n+2)(4n+3)}{6} -2^{2}\frac{n(n+1)(2n+1)}{6} ]=\\ =4\frac{2(2n+1)(n+1)(4n+3)-4n(n+1)(2n+1)}{6} =\\ =4\frac{2(n+1)(2n+1)(4n+3-2n)}{6} =\\ =\frac{8(n+1)(2n+1)(2n+3)}{6} =\frac{4(n+1)(2n+1)(2n+3)}{3}$