Question

Tính tổng: \(S=C^1_{20}+2C^2_{20}+2^2C^3_{20}+...+2^{19}C^{20}_{20}\)

Answer 1

Xét khai triển: 

\(\left(x+1\right)^{20}=C_{20}^0+C_{20}^1x+C_{20}^2x^2+...+C_{20}^{20}x^{20}\)

Chia 2 vế cho x ta được:

\(\dfrac{\left(x+1\right)^{20}}{x}=\dfrac{1}{x}+C_{20}^1+C_{20}^2x+...+C_{20}^{20}.x^{19}\)

Thay \(x=2\)

\(\Rightarrow\dfrac{3^{20}}{2}=\dfrac{1}{2}+C_{20}^1+2C_{20}^2+2^2C_{20}^3+...+2^{19}C_{20}^{20}\)

\(\Rightarrow S=\dfrac{3^{20}-1}{2}\)

Answer 2

`S=C_20 ^1 + 2C_20 ^2 + 2^2 C_20 ^3 +....+2^19 C_20 ^20`

`<=>2S=2C_20 ^1+2^2 C_20 ^2 + 2^3 C_20 + .... + 2^20 C_20 ^20`

`<=>2S=C_20 ^0 +2C_20 ^1+2^2 C_20 ^2 + 2^3 C_20 + .... + 2^20 C_20 ^20 -C_20 ^0`

`<=>2S=(1+2)^20-1`

`<=>2S=3^20-1`

`<=>S=[3^20 -1]/2`