若C0n−4C1n+42C2n−43C3n+⋯+(−1)n4nCnn=−243,则C1n+C2n+⋯+Cnn=( ).A.64B.32C.63D.31
求证:(1)C0n−C2n+C4n−C6n+⋯=(√2)ncosnπ4;(2)C1n−C3n+C5n−C7n+⋯=(√2)nsinnπ4.
1n+1=C0n−12C1n+13C2n+⋯+(−1)n−1nCn−1n+(−1)nn+1Cnn证毕. Cknk+1=Ck+1n+1n+1, 左边=n∑k=0(−1)kn+1Ck+1n+1=−1n+1n∑k=0Ck+1n+1(−1)k+1=−1n+1[(1−1)n+1−1]=1n+1=右边, ∴C0n−12C1n+13C2n−14C3n+⋯+(−1)nn+1Cnn=1...
答案B答案 B解析 由已知条件得 (1+2)n=3n=729,解得 n= 6.2-|||--16-|||-的展开式中不含-|||-2.二项式-|||-×3项的系数之和为Cn1+ C3n+Cn5=C61+ C36+C65=32.A.20 B.24 C. 30 D.36 答案 A解析 由二项式的展开式的通项公式Tk-1=C6(-1)-|||-12-|||-3k令 12- 3k=3,...
sinnθ=C1n(cosn−1θ)sinθ−C3n(cosn−3θ)sin3θ+C5n(cosn−5θ)sin5θ−⋯. 所以tannθ=sinnθcosnθ=C1n(cosn−1θ)sinθ−C3n(cosn−3θ)sin3θ+C5n(cosn−5θ)sin5θ−⋯C0n(cosnθ)−C2n(cosn−2θ)sin2θ+C4n(cosn−4θ)sin4θ−⋯. ...