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

一
用S表示等比数列{an}的前k项和求证：Sk、S2k-Sk、S3k-S2k也成等比数列.
Sk=a1(1-q^k)/(1-q)=[a1/(1-q)]*(1-q^k)
S2k=a1(1-q^2k)/(1-q) S2k-Sk=[a1/(1-q)]*(q^k-q^2k)
S3k=a1(1-q^3k)/(1-q) S3k-S2k=[a1/(1-q)]*(q^2k-q^3k)
(S2k-Sk)/Sk=(q^k-q^2k)/(1-q^k)=q^k
(S3k-S2k)/(S2k-Sk)=(q^2k-q^3k)/(q^k-q^2k)=q^k
Sk、S2k-Sk、S3k-S2k也成等比数列.
二
等比数列
sk=a1+a2+……+ak
s2k-sk=a(k+1)+a(k+2)+……+a2k
因为a(k+1)=a1*q^k,a(k+2)=a2*q^k……a2k=ak*q^k
所以s2k-sk=a(k+1)+a(k+2)+……+a2k
=a1*q^k+a2*q^k+……+ak*q^k
=(a1+a2+……+ak)*q^k
=sk*q^k
同理s3k-s2k=a(2k+1)+a(2k+2)+……+a3k
=a(k+1)*q^k+a(k+2)*q^k+……+a2k*q^k
=(a(k+1)+a(k+2)+……+a2k)*q^k
=(s2k-sk)*q^k
综上所述sk,s2k-sk,s3k-s2k为公比为q^k的等比数

路过 我也想知道WHY