已知等差数列{an}满足∶a3=7,a5+a7=26,{an}的前几项和为sn令bn=2的n次方乘an,求数列前n项和T

∵a5 = a3 + 2d
a7 = a3 + 4d
∴a5+a7= 2a3 + 6d = 26
又∵a3 = 7
∴d = 2
∴a1 = a3 - 2d = 3
∴an = a1 + (n-1)d = 3 + 2(n-1) = 2n + 1
∴bn = (2n+1)×2^n
Tn = 3×2 + 5×2^2 + 7×2^3 + …… + (2n+1)×2^n
2Tn = 3×2^2 + 5×2^3 + …… + (2n-1)×2^n + (2n+1)×2^(n+1)
两式相减,得：
Tn = (2n+1)×2^(n+1) - 3×2 - [2^3 + 2^4 + …… + 2^(n+1)]
= (2n+1)×2^(n+1) - 6 - 8[1-2^(n-1)]/(1-2)
= (2n+1)×2^(n+1) - 6 - 8×2^(n-1) + 8
= (2n+1)×2^(n+1) - 6 - 2×2^(n+1) + 8
= (2n-1)×2^(n+1) + 2

a3=7 a5+a7=26
so,an=6n-11
bn=2^n*an=2^n(6n-11)
Tn=b1+b2+⋯⋯＋bn然后用裂项求和就可以了详细一点的过程可以不....这里学的不是特别好，看过程我会更明白一些，麻烦了Tn=2(6-11)+2^2(6*2-11)+2^3(6*3-11)＋⋯⋯＋2^(n-1)(6(n-1)-...