: ${{S}_{2}}=C_{2011}^{0}+{{2}^{2}}C_{2011}^{2}+…+{{2}^{2010}}C_{2011}^{2010}$

C. $\frac{{{3}^{2011}}+1}{2}$

B. $\frac{{{3}^{211}}-1}{2}$

C. $\frac{{{3}^{2011}}+12}{2}$

D. $\frac{{{3}^{2011}}-1}{2}$

Hướng dẫn

Chọn D

Xét khai triển:

${{(1+x)}^{2011}}=C_{2011}^{0}+xC_{2011}^{1}+{{x}^{2}}C_{2011}^{2}+…+{{x}^{2010}}C_{2011}^{2010}+{{x}^{2011}}C_{2011}^{2011}$

Cho $x=2$ ta có được:

${{3}^{2011}}=C_{2011}^{0}+2.C_{2011}^{1}+{{2}^{2}}C_{2011}^{2}+…+{{2}^{2010}}C_{2011}^{2010}+{{2}^{2011}}C_{2011}^{2011}$ (1)

Cho $x=-2$ ta có được:

$-1=C_{2011}^{0}-2.C_{2011}^{1}+{{2}^{2}}C_{2011}^{2}-…+{{2}^{2010}}C_{2011}^{2010}-{{2}^{2011}}C_{2011}^{2011}$ (2)

Lấy (1) + (2) ta có:

$2\left( C_{2011}^{0}+{{2}^{2}}C_{2011}^{2}+…+{{2}^{2010}}C_{2011}^{2010} \right)={{3}^{2011}}-1$

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