ccf83dy60d的部落格

標題:

Binomial Theorem\\\\\\\\\\\\\\

發問:

Simplify ∑(r=0→n) [1/(r+2)] n_C_r 更新: ANS: [n●2^(n+1) +1]/[(n+1)(n+2)]

最佳解答:

x(1+x)?=∑((r=0→n)n_C_r x??1 Integrate both sides wrt x from 0 to 1 ∫(x=0→1)x(1+x)?dx=∫(x=0→1){∑((r=0→n)n_C_r x??1} dx ∫(x=0→1)x(1+x)?dx=∑((r=0→n)n_C_r ∫(x=0→1) x??1 dx ∫(x=0→1)[(1+x)??1-(1+x)?]dx=∑((r=0→n)n_C_r ∫(x=0→1) x??1 dx [(1+x)??2/(n+2)-(1+x)??1/(n+1)](x=0→1)=∑((r=0→n)n_C_r [x??2/(r+2)](x=0→1) (2??2-1)/(n+2)-(2??1-1)/(n+1)=∑((r=0→n)n_C_r [1/(r+2)] [(n+1)(2??2-1)-(n+2)(2??1-1)]/[(n+1)(n+2)]=∑((r=0→n)n_C_r/(r+2) ∑((r=0→n)n_C_r/(r+2)=[n(2??1)+1]/[(n+1)(n+2)]

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