(2+1/2)^2 + (4+1/4)^2 +...+(2^n+1/2^n)^2 Найти сумму. Тема: алгебраическая и...

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

$S_{n} = 2^2+4^2+...(2^n)^2 + 2*(1+1+1+.+n)+\frac{1}{2}^2 + \frac{1}{4}^2+...+(\frac{1}{2^n})^2 \\\\ S_{1}=2^2+2^4+2^6 + ... (2^n)^2 = \frac{4(4^n - 1 ) }{3} \\ S_{2}=2n \\ S_{3} = \frac{ (-4^{-n}+1)}{3}\\S_{n}=\frac{ 6n- 4^{-n}+4^{n+1} - 3 }{ 3 }$