Summing +summing+summing...

Well Parth Lohomi, you should not rate all your problems at level 5 because not that they will be attractive(as I think that blue level 5 is truly attractive) but that may frustrate the user as after solving the problem, he really get irritated by its simplicity(as he might overthink it.

Well, yeah the problem -

$\displaystyle \sum_{r=0}^{n}{{(-1)}^{r}C(n,r)\frac{1+rln10}{{(1+nln10)}^{r}}}$

$\displaystyle \sum_{r=0}^{n}{C(n,r){\frac{-1}{1+nln10}}^{r} + rln10C(n,r){\frac{-1}{1+nln10}}^{r}}$

The first one is simply binomial expansion and the second one will be transformed to be a binomial expansion as well.

$\displaystyle {(1 - \frac{1}{1+nln10})}^{n} + \sum_{r=1}^{n}{nln10C(n-1,r-1){\frac{-1}{1+nln10}}^{r}}$ [As $C(n,r) = \frac{n}{r}C(n-1,r-1)$ , lower bound of the sum as zero just equal to 0, hence it is changed to 1]

$\displaystyle {(1 - \frac{1}{1+nln10})}^{n} + nln10\frac{-1}{1+nln10}\sum_{r=1}^{n}{C(n-1,r-1)\frac{-1}{1+nln10}]^{r-1}}$

And hence, the second one is also a binomial expansion, so,

$\displaystyle {(1 - \frac{1}{1+nln10})}^{n} + nln10\frac{-1}{1+nln10}{(1 - \frac{1}{1+nln10})}^{n-1}$

$\displaystyle = {(1 - \frac{1}{1+nln10})}^{n-1}(1 - \frac{1}{1+nln10} - \frac{nln10}{1+nln10})$

which is just equal to $\boxed{0}$