Summation of more than one series How can I find $1+3x+6x^2+10x^3+.... $ for $x=6/7$. I have seen that this can be written as a sum of three series $\sum_0^\infty [(n^2/2)x^n+(3/2)nx^n+x^n]$. But I do not know how to proceed further.
 A: Let's give an example of how to calculate one of your terms:  $\sum_0^\infty nx^n$
$$\sum_0^\infty nx^n$ =\sum_1^\infty x^n + \sum_2^\infty x^n +\sum_3^\infty x^n$ + \cdots
$$
(You can see that each term of the form $x^k$ appears in $k$ of these sums, so this is just a rearrangement of the terms in $\sum_0^\infty nx^n$.)
Each of those sums are easier.  For example:
$$
\sum_m^\infty x^n = x^m \sum _0^\infty x^n = \frac{x^m}{1-x}
$$
And these answers are all different by factors of $x$, so we have
$$
\sum_0^\infty nx^n =\frac{1}{1-x} \sum_{m=1}^\infty x^m = \frac{x}{(1-x)^2} 
$$
where the $x$ in the numerator comes from the fact that the sum starts at $m=1$ not $0$.
Tackle the $\sum_0^\infty n^2 x^n$ the same way:
$$\sum_0^\infty n^2 x^n = 
 \sum_1^\infty n x^n + \sum_2^\infty n x^n + \cdots
$$
The answer works out to $$\sum_0^\infty n^2 x^n = \frac{x^2+x}{(1-x)^3}$$
A: Note that you have sum of terms of type $\frac{(n+1)(n+2)}{2}x^n$, so your sum would be:
$$1+(1+2)x+(1+2+3)x^2+(1+2+3+4)x^3+...=\\=(1+x+x^2+x^3+...)+2(x+x^2+x^3+...)+3(x^2+x^3+x^4+...)+...=\\=(1+x+x^2+x^3+...)+2x(1+x+x^2+x^3...)+3x^2(1+x+x^2+x^3+...)+...$$ Since $x=6/7$ we have $1+x+x^2+x^3+...=7$, so the above expression is: 
$$=7(1+2x+3x^2+4x^3+...)$$ Now, calculate $1+2x+3x^2+4x^3+...$ by similar method, and you will obtain result.
A: Hint: Use the fact that a power series can be differentiated term-by-term within its radius of convergence. 
For example:
The power series $\sum_0^\infty x^{n+2}$ is a geometric series that converges to $A(x):={x^2\over 1-x}$ if $|x|<1$. The power series can be differentiated with respect to $x$, term by term, and the result  will converge to $A'(x)$ if $|x|<1$:
$$
\sum_0^\infty(n+2)x^{n+1} = A'(x) = {d\over dx}{x^2\over 1-x}={1\over(1-x)^2}-1\;.
$$
Now do this one more time. The result when $x=6/7$ will differ from your series by a factor of 2.
A: You properly identified that you have to compute $$f(x)=\frac 12\sum_{n=0}^\infty (n^2+3n+2)x^n$$ Rewrite $$n^2+3n+2=\big(n(n-1)+n\big)+3n+2=n(n-1)+4n+2$$  So, expanding $$f(x)=\frac 12\sum_{n=0}^\infty n(n-1)x^n+2\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n$$ $$f(x)=\frac {x^2}2\sum_{n=0}^\infty n(n-1)x^{n-2}+2x\sum_{n=0}^\infty nx^{n-1}+\sum_{n=0}^\infty x^n$$ in which you successively recognize the second and first derivatives of the last summation which is quite well known.
I am sure that you can take from here.
