1、 (1)(1-x)∑(k=1…n)xk-1=1-xn. (2)(x)∑(k=1…n)(1-x)k-1=1-(1-x)n (3)∑(k=1…n)(1-x)k-1=∑(k=1…n)C(n,k)(-x)k-1 (4)积分区间[0,1]: ∑(k=1…n)(1/k)=∑(k=1…n)C(n,k)(-1)k-1(1/k) (5)∑(k=1…n)(1/k)=∫(0,1)[(1-xn)/(1-x)]dx. 2、 |
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来自: toujingshuxue > 《数学》