Formula for the summation of the series $a+2a^2+3a^3+...$ upto nth term?

Question

I need a formula for the summation of the series $a+2a^2+3a^3+...$ upto nth term. Is there any way to simplify it?

Answer 1

You request $\sum_{j=1}^n j a^j$. Then we have \begin{align*} \sum_{j=1}^n j a^j &= a \sum_{j=1}^n j a^{j-1} \\ &= a \sum_{j=1}^n \frac{\mathrm{d}}{\mathrm{d}a} a^j \\ &= a \frac{\mathrm{d}}{\mathrm{d}a} \sum_{j=1}^n a^j \\ &= a \frac{\mathrm{d}}{\mathrm{d}a} \frac{a^{n+1} - a}{a-1} \\ &= a \frac{n a^{n+1} - n a^n - a^n + 1}{(a-1)^2} \\ &= \frac{a^{n+1}(an - n -1) + a}{(a-1)^2} \text{.} \end{align*}

Answer 2

$a+2a^2+3a^3+\dots+na^n=a(1+2a+3a^2+\dots+na^{n-1})=\\=a\dfrac{d}{da}(a+a^2+\dots+a^n)=a\dfrac{d}{da}\left(\dfrac{a-a^{n+1}}{1-a}\right)= \dfrac{a-(n + 1)a^{n+1}+na^{n+2} }{(1-a)^2} .$

Answer 3

Let $a+2a^2+...+na^n=S$, $$aS=a^2 + 2a^3 +3a^4+...+(n-1)a^n+na^{n+1}$$

Then $$aS-S=na^{n+1}-(a^n+a^{n-1}+...+a^2+a)$$

Since $$a+a^2+...+a^n=\frac{a^{n+1}-a}{a-1}$$

$$S=\frac{na^{n+1}- \frac{a^{n+1}-a}{a-1}}{a-1}$$

$$S=\frac{na^{n+2}-(n+1)a^{n+1}+a}{(a-1)^2}$$

Answer 4

$\frac{(a\,*\,(1+na^{(1+n)}\,-\,(1+n)*a^n))}{(-1+a)^2}$

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