用数学归纳法证明1-2^2+3^2-4^2+...+(-1)^(n-1)*n^2=(-1)^(n-1)*n(n-1)/2
\u7528\u6570\u5b66\u5f52\u7eb3\u6cd5\u8bc1\u660e1^2-2^2+3^2-4^2+~+(2n-1)^2-(2n)^2=-n(2n+1)1 n=1\u65f6\uff0c\u5de6\u8fb9=\uff082*1-1\uff09^2-2^2=-3
\u53f3\u8fb9=-1\uff082*1+1\uff09=-3=\u5de6\u8fb9
2 \u5f53n=k\u65f6\uff0c\u5047\u8bbe\u7b49\u5f0f\u6210\u7acb\uff0c\u5de6\u8fb9=\u53f3\u8fb9\uff0c
\u5373 1^2-2^2+3^2-4^2+~+(2k-1)^2-(2k)^2=-k(2k+1)
\u90a3\u4e48\u5f53n=k+1\u65f6\uff0c\u5de6\u8fb9=1^2-2^2+3^2-4^2+~+(2k-1)^2-(2k)^2+(2k+1)^2-(2k+2)^2=-k(2k+1)+(2k+1)^2-(2k+2)^2=-2k^2-k+4k^2+4k+1-4k^2-8k-4=-2k^2-5k-3=-(k+1)(2k+3)=\u53f3\u8fb9
\u5373 \u5f53n=k+1\u65f6\u7b49\u5f0f\u4e5f\u6210\u7acb\u3002
3 \u547d\u9898\u5f97\u8bc1
1 n=2\u6ee1\u8db3
2\u8bben=k-1\u6ee1\u8db3\uff0c\u8bc1n=k\u6ee1\u8db3\u5373\u53ef\uff08\u76f8\u51cf\u5c0f\u4e8e\u4e00\uff0c\u67092^\uff08k-1\uff09\u9879\uff09
先证明一项和成立
1=(-1)^0*1(1+1)/2=1
假设对于n项和成立
1-2^2+3^2-4^2+...+(-1)^(n-1)*n^2=(-1)^(n-1)*n*(n+1)/2
n+1项和
1-2^2+3^2-4^2+...+(-1)^(n-1)*n^2+(-1)^(n)*(n+1)^2
=(-1)^(n-1)*n*(n+1)/2+(-1)^(n)*(n+1)^2
=(-1)^(n)*(n+1)^2-(-1)^(n)*n*(n+1)/2
=(-1)^(n)*[(n+1)^2-n*(n+1)/2]
=(-1)^(n)*[(n+1)(n+1-n/2)]
=(-1)^(n)*(n+1)(n+2)/2
该等式对于n+1项和也成立，由数学归纳法证得1-2^2+3^2-4^2+...+(-1)^(n-1)*n^2=(-1)^(n-1)*n*(n+1)/2对于所有的正整数n成立
lz题目好像有错啊
绛旓細瑙ｉ杩囩▼濡備笅锛
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