数列求和用分组求和及并项法求和Sn=1^2-2^2+3^2-4^2+…+(-1)^(n-1)·n^2
数列求和用分组求和及并项法求和Sn=1^2-2^2+3^2-4^2+…+(-1)^(n-1)·n^2
数列求和用分组求和及并项法求和Sn=1^2-2^2+3^2-4^2+…+(-1)^(n-1)·n^2
数列求和用分组求和及并项法求和Sn=1^2-2^2+3^2-4^2+…+(-1)^(n-1)·n^2
∵Sn=1²-2²+3²-4²+…+(-1)^(n-1)·n²
∴当n是奇数时:
Sn=1²-2²+3²-4²+5²-6²+7²-……+n²
=[1²+2²+3²+4²+5²+…+n²]-2×[2²+4²+6²+8²+…+(n-1)²]
=n(n+1)(2n+1)/6-8×[1²+2²+3²+4²+…+((n-1)/2)²]
=n(n+1)(2n+1)/6-8×[(n-1)/2]×[(n-1)/2+1]×[(n-1)+1]/6
=n(n+1)(2n+1)/6-2×(n-1)(n+1)n/6
=[n(n+1)/6]×[(2n+1)-2×(n-1)]
=[n(n+1)/6]×[1+2]
=n(n+1)/2
∴当n是偶数时:
Sn=1²-2²+3²-4²+5²-6²+7²-……-n²
=[1²+2²+3²+4²+5²+…+n²]-2×[2²+4²+6²+8²+…+n²]
=n(n+1)(2n+1)/6-8×[1²+2²+3²+4²+…+(n/2)²]
=n(n+1)(2n+1)/6-8×[n/2]×[n/2+1]×[n+1]/6
=n(n+1)(2n+1)/6-2×n(n+2)(n+1)/6
=[n(n+1)/6]×[(2n+1)-2×(n+2)]
=[n(n+1)/6]×[1-4]
=-n(n+1)/2
两式合并,得:
1²-2²+3²-4²+5²-6²+7²-……+n²
=[(-1)^(n+1)]×n(n+1)/2
说明:
运用了公式1²+2²+3²+4²+5²+6²+7²+……+n²=n(n+1)(2n+1)/6