Tính S= C0 2011 + 2^2 C2 2011 + ... + 2^2010 C2010 2011

* Hướng dẫn giải

* Xét khai triển:

${(1+x)}^{2011}=C_{2011}^0+x{C}_{2011}^1+x^2{C}_{2011}^2+\mathrm{...}+x^{2010}{C}_{2011}^{2010}+x^{2011}{C}_{2011}^{2011}$

* Cho x= 2  ta có được:

$3^{2011}=C_{2011}^0+2.C_{2011}^1+2^2{C}_{2011}^2+\mathrm{...}+2^{2010}{C}_{2011}^{2010}+2^{2011}{C}_{2011}^{2011}$ (1)

* Cho x= -2 ta có được:

$-1=C_{2011}^0-2.C_{2011}^1+2^2{C}_{2011}^2-\mathrm{...}+2^{2010}{C}_{2011}^{2010}-2^{2011}{C}_{2011}^{2011}$  (2)

* Lấy (1) + (2) ta có:

$2\left(C_{2011}^0+2^2{C}_{2011}^2+\mathrm{...}+2^{2010}{C}_{2011}^{2010}\right)=3^{2011}-1$

Suy ra:$S=C_{2011}^0+2^2{C}_{2011}^2+\mathrm{...}+2^{2010}{C}_{2011}^{2010}=\frac{3^{2011}-1}{2}$.

Chọn đáp án D