Đáp án: $0$
Giải thích các bước giải:
Ta có:
$G=2^2+2^4+2^6+2^8+...+2^{94}+2^{96}$
$\to G=(2^2+2^6+2^{10}+...+2^{94})+(2^4+2^8+...+2^{96})$
$\to G=((2^2+2^6)+(2^{10}+2^{14})+...+(2^{90}+2^{94}))+((2^4+2^8)+...+(2^{92}+2^{96}))$
$\to G=(2^2(1+2^4)+2^{10}(1+2^4)+...+2^{90}(1+2^4))+(2^4(1+2^4)+...+2^{92}(1+2^4))$
$\to G=(1+2^4)(2^2+2^{10}+...+2^{90})+(1+2^4)(2^4+...+2^{92})$
$\to G=(1+2^4)((2^2+2^{10}+...+2^{90})+(2^4+...+2^{92}))$
$\to G=17((2^2+2^{10}+...+2^{90})+(2^4+...+2^{92}))$
$\to G\quad\vdots\quad 17$
$\to G$ chia $17$ dư $0$
