题目

数列bn=2^n(4n-3),求Sn

提问时间：2021-03-13

答案

bn=2^n(4n-3)
= 8[n(2^(n-1))] - 3.2^n
consider
(x^(n+1)-1)/(x-1) = 1+x+x^2+..+x^(n)
[(x^(n+1)-1)/(x-1)]' = 1+2x+3x^2+...+nx^(n-1)
1+2x+3x^2+...+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)^2
put x=2
1+2(2)+3(2)^2+...+n(2^(n-1)) = (n)2^(n+1)-(n+1)2^n +1
bn=2^n(4n-3)
= 8[n(2^(n-1))] - 3.2^n
summation bn
= 8[(n)2^(n+1)-(n+1)2^n +1] - 6(2^n-1)
= 8[(n)2^(n+1)-(n+1)2^n +1] - 3(2^(n+1)) +6
=(8n-3).2^(n+1) - 8(n+1)2^n +14