Tính tổng

- Hướng dẫn giải

Theo nhị thức Newton ta có:

${(3+x)}^{2015}=\text{}{C}_{2015}^0.3^{2015}+\text{\hspace{0.17em}\hspace{0.17em}}{C}_{2015}^1{.3}^{2014}.x+\text{}{C}_{2015}^2{.3}^{2013}.x^2+\mathrm{....}+C_{2015}^{2014}\mathrm{.3.}{x}^{2014}+\text{}{C}_{2015}^{2015}.x^{2015}$

Thay x = -1 ta được:

${(3-1)}^{2015}=\text{}{C}_{2015}^0.3^{2015}-\text{\hspace{0.17em}\hspace{0.17em}}{C}_{2015}^1{.3}^{2014}+\text{}{C}_{2015}^2{.3}^{2013}-\mathrm{....}+C_{2015}^{2014}.3-\text{}{C}_{2015}^{2015}$

Suy ra,  $S = 2^{2015}$

 Ta chọn đáp án A