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等差数列an依次每项k之和仍为等差数列,其公差为原公差的k^2倍,即数列Sk,S2k-Sk,S3k-S2k也为等差数列
怎么证
人气:411 ℃ 时间:2020-02-05 09:58:09
解答
Sk=ka1+k(k-1)d/2S2k=2ka1+2k(2k-1)d/2S3k=3ka1+3k(3k-1)d/2S2k-Sk=ka1+k(3k-1)d/2S3k-S2k=ka1+k(5k-1)d/2(S2k-Sk)-Sk=k^2*d(S3k-S2k)-(S2k-Sk)=k^2*d所以等差数列an依次每项k之和仍为等差数列,其公差为原公差的k^...
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