求证1²-2²+3²-4²+……+(2n-1)²-(2n)²=-n(2n+1)
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![求证1²-2²+3²-4²+……+(2n-1)²-(2n)²=-n(2n+1)](/uploads/image/z/2590344-0-4.jpg?t=%E6%B1%82%E8%AF%811%26%23178%3B%EF%BC%8D2%26%23178%3B%EF%BC%8B3%26%23178%3B%EF%BC%8D4%26%23178%3B%2B%E2%80%A6%E2%80%A6%2B%282n-1%29%26%23178%3B-%282n%29%26%23178%3B%3D-n%282n%EF%BC%8B1%29)
求证1²-2²+3²-4²+……+(2n-1)²-(2n)²=-n(2n+1)
求证1²-2²+3²-4²+……+(2n-1)²-(2n)²=-n(2n+1)
求证1²-2²+3²-4²+……+(2n-1)²-(2n)²=-n(2n+1)
1 n=1时,左边=(2*1-1)^2-2^2=-3
右边=-1(2*1+1)=-3=左边
2 当n=k时,假设等式成立,左边=右边,
即 1^2-2^2+3^2-4^2+~+(2k-1)^2-(2k)^2=-k(2k+1)
那么当n=k+1时,左边=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)=右边
即 当n=k+1时等式也成立.
3 命题得证=
平方差公式就可以化成-(3+7+11+……+4n-1)再用等差数列求和公式((首项加尾项)*项数除二)就得到结果