It Doesn't Telescope

Note that $\begin{aligned} A(x) & = \; \sum_{n=1}^\infty \left(\sum_{j=1}^n (n+1-j)j^2\right)x^n \; = \; \tfrac1{12}\sum_{n=1}^\infty n(n+1)^2(n+2)x^n \\ & = \tfrac1{12}\sum_{n=1}^\infty n(n+1)(n+2)(n+3)x^n - \tfrac16\sum_{n=1}^\infty n(n+1)(n+2)x^n \\ & = \tfrac{1}{12} \times 24x(1-x)^{-5} - \tfrac16 \times 6x(1-x)^{-4} \; = \; \frac{x(1+x)}{(1-x)^5} \end{aligned}$ for all $-1 < x < 1$ , using standard formulae for the Binomial expansions of negative powers of $1-x$ . Thus $A\big(\tfrac{1}{10}\big) = \tfrac{11000}{59049}$ , making the answer $\boxed{70049}$ .

Although I'm not actually sure if the series telescopes, we can solve this problem fairly easily by iterating through several other progressively simpler series:

Let $A=\frac{1^{2}}{10}+\frac{1^{2}+2^{2}}{10^{2}}+\frac{1^{2}+2^{2}+3^{2}}{10^{3}}+\ldots$ .

Then $A+\frac{A}{10}+\frac{A}{10^{2}}+ \ldots = S \Rightarrow \left( \frac{10}{9} \right)A=S$

Let $B=\frac{1^{2}}{10}+\frac{2^{2}}{10^{2}}+\frac{3^{2}}{10^{3}}+ \ldots$

We can see that $A-\frac{A}{10}=B \Rightarrow A=\left( \frac{10}{9} \right)B \Rightarrow \left( \frac{10}{9} \right)^{2}B=S$

Let $C=\frac{1}{10}+\frac{3}{10^{2}}+\frac{5}{10^{3}}+\frac{7}{10^{4}} + \ldots$

We can see that $B-\frac{B}{10}=C \Rightarrow B=\left( \frac{10}{9} \right)C \Rightarrow \left( \frac{10}{9} \right)^{3}C=S$

Let $D=\frac{2}{10^{2}}+\frac{2}{10^{3}}+\frac{2}{10^{4}}+\frac{2}{10^{5}}+ \ldots = \frac{2}{90}$

We can see that $C-\frac{C}{10}=\frac{1}{10}+D =\frac{11}{90} \Rightarrow C=\frac{11}{9^{2}} \Rightarrow S=\frac{11 \times 10^{3}}{9^{5}}=\frac{11000}{59049}$

We see from the exponential form of the fraction that $11000$ and $59049$ are co-prime, and so our answer is $11000+59049=\boxed{70049}$

Note that this only holds true if $S,A,B,C,D$ are convergent. Since the system implies that $S>A>B>C>D$ , it will suffice to show that $S$ is convergent.

$\frac{n \times 1^{2}+(n-1) \times 2^{2}+(n-2) \times 3^{2} + \ldots + n^{2}}{10^{n}} \leq \frac{(n+1)n^{3}}{2 \times 10^{n}}$ for all $n=1,2,3, \ldots$

Since the infinite series corresponding to $a_{n}=\frac{(n+1)n^{3}}{2 \times 10^{n}}$ converges by the ratio test, we can conclude that $S,A,B,C,D$ all converge.

First we find the sum $n\times 1^2+(n-1)\times 2^2+(n-2)\times 3^2+\cdots+1\times n^2=\sum\limits_{k=1}^n (n-k+1)k^2$ . To do this, we start with the sum $\sum\limits_{k=1}^n(n-k+1)((k+1)^3-k^3)=1^3+2^3+\cdots+n^3+n(n+1)(n+2)=\frac14 n(n+1)(n^2+5n+8)$ . On the other hand $\sum\limits_{k=1}^n (n-k+1)(3k^2+3k+1)=3\sum\limits_{k=1}^n (n-k+1)k^2+3\sum\limits_{k=1}^n (n-k+1)k+\sum\limits_{k=1}^n (n-k+1)$ . So, we obtain $\sum_{k=1}^n (n-k+1)k^2=\frac13\Big(\frac14 n(n+1)(n^2+5n+8)-\frac12n(n+1)(n+2)-\frac12n(n+1)\Big)=\frac1{12}n(n+1)^2(n+2)$

Now, we consider the power series $\frac1{12}\sum_{n=1}^\infty n(n+1)^2(n+2)x^n.$ Its radius of convergence is $1$ . We use derivative $\begin{aligned}\frac1{12}\sum_{n=1}^\infty n(n+1)(x^{n+2})'' &=\:\frac1{12}\Big(\sum_{n=1}^\infty n(n+1)x^{n+2}\Big)''=\frac1{12}\Big(x^3\sum_{n=1}^\infty n(n+1)x^{n-1}\Big)''=\frac1{12}\Big(x^3\sum_{n=1}^\infty (x^{n+1})''\Big)''\\ &=\:\frac1{12}\Big(x^3\Big(\sum_{n=1}^\infty x^{n+1}\Big)''\Big)''=\frac1{12}\Big(x^3\Big(x^2\sum_{n=1}^\infty x^{n-1}\Big)''\Big)''=\frac1{12}\Big(x^3\Big(\frac{x^2}{1-x}\Big)''\Big)''\\ &=\:\frac1{12}\Big(\frac{2x^3}{(1-x)^3}\Big)''=\frac{x(x+1)}{(1-x)^5}.\end{aligned}$ Replacing $x=\dfrac1{10}$ in the last expression yields result $\dfrac{11000}{59049}$ , meaning $11000+59049=70049.$