若等比数列前n项,前2n项,前3n项的和分别为sn s2n s3n 求证sn∧2+s2n∧2=sn(s2n+s3n)
人气:379 ℃ 时间:2020-04-11 16:14:05
解答
an = a1q^(n-1)Sn = a1(q^n-1)/(q-1)(Sn)^2 + (S(2n))^2= [a1(q^n-1)/(q-1)]^2 +[a1(q^(2n)-1)/(q-1)]^2= [a1/(q-1)]^2 .[ q^(4n) -q^(2n) - 2q^n + 2]Sn[S(2n) + S(3n) ]=[a1(q^n-1)/(q-1)] .[ a1(q^(2n)-1)/(q-1) ...
推荐
- 已知等比数列的前n项,前2n项,前3n项.求证Sn^2+S2n^2=Sn(S2n+S3n)
- 等比数列前n项,前2n项,前3n项的和分别是Sn,S2n,S3n,求证Sn,S2n-Sn,S3n-S2n也成等比
- 等比数列前n项,前2n项,前3n项的和分别是Sn,S2n,S3n,求证Sn,S2n-Sn,S3n-S2n也成等比数列
- 已知等比数列{an}的前n项和Sn=54,前2n项和S2n=60,则前3n项和S3n=( ) A.64 B.66 C.6023 D.6623
- 一个等比数列{an}的前n项和Sn=48,前2n项S 2n=60,则前3n项和S 3n=( )
- “咽”有几种读音?
- 612-375+275+(388+286)简便算法
- 肥皂英文单词怎么说?
猜你喜欢