nhị thức niuton 3

Question

Tính tổng sau :
a/ $S_1=C^6_{11}+C^7_{11}+C^8_{11}+C^9_{11}+C^{11}_{11}$.
b/ $S_2=3^{16}C^0_{16}-3^{15}C^1_{16}+3^{14}C^2_{16}-.....+C^{16}_{16}$

score 1 · Answer 1

Ta có $(1+1)^{11} = C_{11}^0+C_{11}^1+C_{11}^2+...+C_{11}^6 +...+ C_{11}^{10} + C_{11}^{11}$

$(1+1)^5 = C_{11}^0 +C_{11}^1+...+C_{11}^5$

Trừ 2 vế cho nhau $(1+1)^{11} -(1+1)^5 = C_{11}^6 +...+ C_{11}^{10} + C_{11}^{11} = 2^{11}-2^5=2016$

b) Xét $(3-1)^{16} = 3^{16}C_{16}^0 - 3^{15}C_{16}^1+3^{14}C_{16}^2 -...+C_{16}^0 = 2^{16}$

1 Đáp án