求和:1-(1/2)C(n,1)+(1/3)C(n,2)-.....+((-1)^n)/(n+1))C(n,n)=?

2020-06-22 9:04 pm

回答 (1)

2020-06-22 10:14 pm
✔ 最佳答案
1-(1/2)C(n,1)+(1/3)C(n,2)-.....+((-1)^n)/(n+1))C(n,n)
    = C(n+1,1)/(n+1) - C(n+1,2)/(n+1) + C(n+1,3)/(n+1) +
                                          ...+(-1)^nC(n+1,n+1)/(n+1)
    = 1/(n+1) -C(n+1,0)/(n+1)
              + C(n+1,1)/(n+1) - C(n+1,2)/(n+1)
                                     + ...+(-1)^nC(n+1,n+1)/(n+1)

    = 1/(n+1) - [1/(n+1)]{C(n+1,0)-C(n+1,1)+...+(-1)^(n+1)C(n+1,n+1)}
    = 1/(n+1) - [1/(n+1)](1-1)^(n+1)
    = 1/(n+1)
    


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