Find the sum of $1.2+2.3x+3.4x^2...\infty$ where $|x|<1$
I have got the $\mathbb{n^{th}\quad term \quad= n(n+1)(x^{n-1})}$
Then I tried finding $\sum\limits_{k=1}^\infty n(n+1)(x^{n-1})$ but it's in product form so couldn't.
How do I continue from here?
On
Suppose you know what $$f(x)=1+2x+3x^2+4x^3+\cdots$$ is. Then $$f'(x)=2+2\times 3x+3\times 4x^2+4\times 5x^3+\cdots.$$ Exactly what you want!
Perhaps you already know what $f(x)$ is, or maybe now you have a clue about how to find it.
On
If $f(x) =\sum_{n=1}^{\infty} n(n+1)x^n $ then
$\begin{array}\\ xf(x) &=x\sum_{n=1}^{\infty} n(n+1)x^n\\ &=\sum_{n=1}^{\infty} n(n+1)x^{n+1}\\ &=\sum_{n=2}^{\infty} n(n-1)x^{n}\\ &=\sum_{n=1}^{\infty} n(n-1)x^{n}\\ \end{array} $
so $f(x)-xf(x) =\sum_{n=1}^{\infty} (n(n+1)-n(n-1))x^{n} =2\sum_{n=1}^{\infty} nx^{n} $ or $f(x) =\dfrac{2}{1-x}\sum_{n=1}^{\infty} nx^{n} $.
Similarly, if $g(x) =\sum_{n=1}^{\infty} nx^{n} $ then $xg(x) =\sum_{n=1}^{\infty} nx^{n+1} =\sum_{n=2}^{\infty} (n-1)x^{n} =\sum_{n=1}^{\infty} (n-1)x^{n} $ so $g(x)-xg(x) =\sum_{n=1}^{\infty} (n-(n-1))x^{n} =\sum_{n=1}^{\infty} x^{n} =\dfrac{x}{1-x} $ and $g(x) =\dfrac{x}{(1-x)^2} $.
Therefore $f(x) =\dfrac{2}{1-x}g(x) =\dfrac{2x}{(1-x)^3} $.
Your sum is $\dfrac{f(x)}{x} =\dfrac{2}{(1-x)^3} $.
On
Notice the following recurrence:
$$t_n-2t_{n-1}+t_{n-2}=(n+1)(n+2)-2n(n+1)+(n-1)n=2.$$
Then summation from $0$ to $n$ with a factor $x^k$ yields (with $t_{-1}=t_{-2}=0$),
$$S_n-2S_{n-1}+S_{n-2}=(1-2x+x^2)S_n=2\frac{1-x^{n+1}}{1-x}.$$
Hence
$$S_\infty=\frac2{(1-x)^3}.$$
This is validated bythe following check:
$$\frac2{0.99^3}=2.06\,12\,20\,30\,42\cdots$$
On
Note that $$\begin{align} (1-x)^{-n} &=\sum_{r=0}^n\binom {-n}r (-x)^r\\ &=\sum_{r=0}^n (-1)^r \binom {n+r-1}r(-1)^r x^r\\ &=\sum_{r=0}^n \binom {n+r-1}r x^r\end{align}$$ The required summation is $$\begin{align} \sum_{r=1}^\infty r(r+1)x^{r-1} &=2\sum_{r=1}^\infty \binom {r+1}2 x^{r-1}\\ &=2\sum_{r=0}^\infty \binom {r+2}2 x^r\\ &=2\sum_{r=0}^\infty \binom {r+2}r x^r\\ &=2\sum_{r=0}^\infty \binom {3+r-1}r x^r\\ &=\color{red}{\frac 2{(1-x)^3}}\end{align}$$
Let $S(x)$ be our sum.
Thus, $$\int\limits_0^x\left(\int\limits_0^xS(x)dx\right)dx=x^2+x^3+...=\frac{x^2}{1-x},$$ which says $$S(x)=\left(\frac{x^2}{1-x}\right)''=\frac{2}{(1-x)^3}.$$
For the solution without calculus you can use the following idea. $$1+2x+3x^2+...=(1+x+x^2+...)+(x+x^2+...)+...=$$ $$=\frac{1}{1-x}+\frac{x}{1-x}=...=\frac{1}{1-x}(1+x+x^2+...)=\frac{1}{(1-x)^2}$$ and $$1\cdot2+2\cdot3x+3\cdot4x^2+...=2\left((1+2x+3x^2+...)+(x+2x^2+3x^3+...)+...\right)=$$ $$=2\left(\frac{1}{(1-x)^2}+\frac{x}{(1-x)^2}+...\right)=\frac{2}{(1-x)^3}.$$ Because $$1\cdot2=2\cdot(1);$$ $$2\cdot3=2(2+1);$$ $$3\cdot4=2(3+2+1);$$ $$.$$ $$.$$ $$.$$ $$n(n+1)=2(n+(n-1)+...+1)=2\cdot\frac{n(n+1)}{2}.$$