已知数列｛an｝的前n项和为sn,且sn=2n^2+n,n是正整数,又an=4log（2）bn+3（

已知数列｛an｝的前n项和为sn,且sn=2n^2+n,n是正整数,又an=4log（2）bn+3（

(1)Sn= 2n^2+n (1)S(n-1) = 2(n-1)^2 + (n-1) (2)(1) -(2)an= 2n^2+n - 2(n-1)^2 - (n-1) = 4n-1an=4log(2)bn+34n-1 = 4log(2)bn+3log(2)bn = n-1bn = 2^(n-1)(2)anbn = (4n-1)(2^(n-1)) = 4[n2^(n-1)] - 2^(n-1)consider[x^(n+1)-1]/(x-1) = 1+x+x^2+...+x^n[(x^(n+1)-1)/(x-1)]' = 1+2x+..+nx^(n-1)1+2x+..+nx^(n-1) = [ (x-1)(n+1)x^n) -(x^(n+1)-1) ]/(x-1)^2 = [nx^(n+1)-(n+1)x^n+1]/(x-1)^2put x=21+2(2)+3(2)^2+..+n(2)^(n-1)=n2^(n+1)- (n+1)2^n +1summation anbn=summation {4[n2^(n-1)] - 2^(n-1)}= 4(n2^(n+1)- (n+1)2^n +1) - (2^n-1)=(4n).2^(n+1) - (4n+5).2^n +5======以下答案可供参考======供参考答案1:an*bn=n*2^(n+1)-2^(n-1)后面2^(n-1)直接套等比数列求和公式令cn=n*2^(n+1)故1/2cn=n*2^nTn=c1+c2+c3……+cn=1*4+2*8+3*16+4*32……+(n-1)*2^n+n*2^(n+1)1/2Tn=…… =1*2+2*4+3*8+…… +n*2^n（看对齐的2^k的系数都相差1）Tn-1/2Tn=-(2+4+8……+2^n)+n*2^(n+1)就可以求了主要思想就是错位相减得到一个等比数列求和就行了供参考答案2:(1-q)sn

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