How many triplet $(a,b,c)$ are there such that $0\le a,b,c \le L$ and $\sum_{x=0}^{\infty}{(ax^{2}+bx+c)\times (\frac{1}{10})^{x+1}} = \frac{p}{q}$ , for given $p,q,L$. Inside Math: Suppose, $A = \sum_{x=0}^{\infty}{ax^2\times (\frac{1}{10})^{x+1}}$ $B = \sum_{x=0}^{\infty}{bx\times (\frac{1}{10})^{x+1}}$ $C = \sum_{x=0}^{\infty}{c\times (\frac{1}{10})^{x+1}}$ Hence, $\sum_{x=0}^{\infty}{(ax^{2}+bx+c)\times (\frac{1}{10})^{x+1}} = A+B+C$ Now, we know(Actually we don’t know this, we have to prove this. We’ll try to prove this in some other section) that: $\sum_{x=0}^{\infty}{x^{2}r^{x+1}} = \frac{r^{2}(r+1)}{(r-1)^{3}} \tag{1}$ $\sum_{x=0}^{\infty}{xr^{x+1}} = \frac{r^{2}}{(r-1)^{2}} \tag{2}$