Question

2²+6²+...+(4n+2)² = ?

Answer 1

cum ultimul termen e (4n+2)^2 =>
termenul general 
(4n+2)^2 =[2*(2n+1)]^2 =4*(2n+1)^2 =>

Suma devine : S=4*(1^2+3^2 +5^2+...+(2n+1)^2 )

aici aplici doar formula patratelor nr impare:

    1^2+3^2+...+            (2p-1)^2        =p*(2p-1)*(2p+1)/3
in cazul tau ai
    (1^2+3^2 +5^2+...      +(2n+1)^2)  , adica p=n+1, inlocuiesti in formula pe p cu n+1 si ai :
=> S=4* (n+1)*(2n+1)*(2n+3)/3
        =4/3 *(n+1)*(2n+1)*(2n+3)
        

Answer 2

$\displaystyle 2^2+6^2+...+(4n+2)^2 \\ \\ \sum\limits ^{n}_{k=0} (4k+2)^2=\sum\limits_{k=0}^n (16k^2+16k+4)=\sum\limits_{k=0}^n16k^2+\sum\limits_{k=0}^n16k+4(n+1)= \\ \\ =16\cdot \sum\limits_{k=0}^nk^2+16\cdot\sum\limits_{k=0}^nk+4n+4=\\ \\=16 \cdot \frac{n(n+1)(2n+1)}{6} +16 \cdot \frac{n(n+1)}{2} +4n+4= \\ \\ = 8 \cdot \frac{n(n+1)(2n+1)}{3} +8 \cdot n(n+1)+4n+4=\\ \\ =8 \cdot \frac{(n^2+n)(2n+1)}{3} +8n^2+8n+4n+4= \\ \\ =\frac{16n ^3+24n^2+8n+24n^2+24n+12n+12}{3} =$
$\displaystyle =\frac{16n^3+48n^2+44n+12}{3} = \frac{4(4n^3+12n^2+11n+3)}{3} = \\ \\ = \frac{4}{3} (4n^3+12n^2+11n+3)= \frac{4}{3} (n+1)(2n+1)(2n+3)$