Tính tổng Sn=1+2a+3a^2+4a^3+....+(n+1)a^n (a khác 1 là số cho trước)

Trả lời:

Nếu a = 0 thì S = 1.

Nếu $a\ne 1$ thì ta có:

$a{S}_n=a+2a^2+3a^3+4a^4+\mathrm{...}+\left(n+1\right)a^{n+1}$

$\Rightarrow S_n-a{S}_n=1+a+a^2+a^3+\mathrm{...}+a^n-\left(n+1\right)a^{n+1}$

$\Rightarrow S_n\left(1-a\right)=\frac{a^{n+1}-1}{a-1}-\left(n+1\right)a^{n+1}$

$\Rightarrow S_n=\frac{1}{1-a}\left[\frac{a^{n+1}-1}{a-1}-\left(n+1\right)a^{n+1}\right]$

$\Rightarrow S_n=\frac{1}{1-a}\left[\frac{a^{n+1}-1-\left(n+1\right)a^{n+1}\left(a-1\right)}{a-1}\right]$

$\Rightarrow S_n=\frac{\left(n+1\right)a^{n+2}-\left(n+2\right)a^{n+1}+1}{{\left(1-a\right)}^2}$

Đáp án cần chọn là: A