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例 求数列 x, 2x2,3x3, … nxn,… 的前n项和. 解:⑴当x=0时 Sn=0⑵当x=1时 Sn=1+2+3+…+ n=n(n+1)/2⑶当x≠1时Sn=x+ 2x2+3x3+ … + nxn ①xSn= x2 +2x3+3x4… + (n-1)xn +nxn +1 ②①-②得:(1-x)Sn=x+ x2+x3+ … +xn - nxn +1 化简得: Sn =x(1- xn )/(1-x) 2 - nxn +1 /(1-x)0 (x=0)综合⑴⑵⑶得 Sn= n(n+1)/2 (x=1)x(1- xn )/(1-x) 2 - nxn +1 /(1-x)(x≠1)小结 1:"错项相减法"求和,常应用于型如{anbn}的数列求和,其中{an}为等差数列, {bn} 为等比数列.答案吗．．．自己对着看咯 |
