JEE Advanced Sequence and Series 2

$S=\displaystyle\sum _{ n=1 }^{ 10 }{ \frac { 15^{ n } }{ { (5^{ n }-3^{ n })(5^{ n+1 }-3^{ n+1 }) } } } \\ S=\displaystyle\frac { 1 }{ 2 } \sum _{ n=1 }^{ 10 }{ \frac { 3^{ n } }{ { (5^{ n }-3^{ n }) } } } -\frac { { 3 }^{ n+1 } }{ (5^{ n+1 }-3^{ n+1 }) } \quad \\ (trying\quad to\quad maintain\quad symmetry\quad and\quad trying\quad to\quad \\ cancel\quad out\quad subsequent\quad terms\quad gives\quad you\quad that)\\ S=\displaystyle\frac { 1 }{ 2 } (\frac { 3 }{ 2 } -\frac { { 3 }^{ 11 } }{ (5^{ 11 }-3^{ 11 }) } )\\ \left\lfloor 10000S \right\rfloor =\boxed { \boxed { 7481 } }$

Is there any other way of doing the last step? One that doesn't involve calculator (seeing how this is a JEE problem)?

Nice problem, @Pranjal Jain

But there is one problem, a little calculation we have to do.

$S = \sum_{n=1}^{10}{\frac{{15}^{n}}{({5}^{n} - {3}^{n})({5}^{n+1} - {3}^{n+1})}}$

We can split this into -

$\sum_{n=1}^{10}{\frac{A}{{5}^{n} - {3}^{n}} + \frac{B}{{5}^{n+1} - {3}^{n+1}}}$

We want ${5}^{n}{3}^{n}$ in the numerator, And easiest would be ${{5}^{n} - {3}^{n}}^{2}$ . So, some bashing gives you-

$\sum_{n=1}^{10}{\frac{A}{{5}^{n} - {3}^{n}} + \frac{{5}^{n} - {3}^{n}}{{5}^{n+1} - {3}^{n+1}}}$

$A$ would be ${5}^{n-1} - {3}^{n-1}$ to cancel out ${5}^{2n} - {3}^{2n}$

$\sum_{n=1}^{10}{\frac{{5}^{n-1} - {3}^{n-1}}{{5}^{n} - {3}^{n}} + \frac{{5}^{n} - {3}^{n}}{{5}^{n+1} - {3}^{n+1}}}$

But here we get an extra $-({5}^{n-1}{3}^{n+1} + {5}^{n+1}{3}^{n-1}) = -{5}^{n}{3}^{n}(\frac{5}{3} + \frac{3}{5})$

So, in the numerator, we finally get -

${5}^{n}{3}^{n}(2 - \frac{5}{3} - \frac{3}{5})$ and $(2 - \frac{5}{3} - \frac{3}{5})$ comes out of the sum.

$(2 - \frac{5}{3} - \frac{3}{5})\sum_{n=1}^{10}{\frac{{5}^{n-1} - {3}^{n-1}}{{5}^{n} - {3}^{n}} + \frac{{5}^{n} - {3}^{n}}{{5}^{n+1} - {3}^{n+1}}}$

The sum will be a telescoping one, so, $\frac{{5}^{10} - {3}^{10}}{{5}^{11} - {3}^{11}}$ . Hence, the answer becomes $\boxed{(2 - \frac{5}{3} - \frac{3}{5})(\frac{{5}^{10} - {3}^{10}}{{5}^{11} - {3}^{11}})}$

We can express the nth term as Tn = 1/2[ 3^n/(5^n - 3^n) - 3^(n+1)/{ 5^(n+1) - 3^(n+1)} ] Therefore summation will be S = 1/2[ 3/2 - 3^11/(5^11 - 3^11) ]