Given,

${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} +$ ...... upto 10 terms

$= {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3}$ + ..... upto 10 terms)

Now,

L.H.S. :-

${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} +$ ...... upto 10 terms

$= 1\,.\,1{C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} +$ ..... upto 10 terms

$= \sum\limits_{r = 1}^{10} {r\,.\,(2r - 1){}^{10}{C_r}}$

$= \sum\limits_{r = 1}^{10} {(2{r^2} - r)\,.\,{}^{10}{C_r}}$

$= 2\,.\,\sum\limits_{r = 1}^{10} {{r^2}\,.\,{}^{10}{C_r} - \sum\limits_{r = 1}^{10} {r\,.\,{}^n{C_r}} }$

[We know, $\sum\limits_{r = 1}^n {r\,.\,{}^n{C_r} = n\,.\,{2^{n - 1}}}$

and $\sum\limits_{r = 1}^n {{r^2}\,.\,{}^n{C_r} = \sum\limits_{r = 1}^n {(r\,.\,{}^n{C_r})\,.\,r} }$

$= \sum\limits_{r = 1}^n {\left( {r\,.\,{n \over r}\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r}$

$= \sum\limits_{r = 1}^n {\left( {n\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r}$

$= n\sum\limits_{r = 1}^n {(r - 1 + 1){}^{n - 1}{C_{r - 1}}}$

$= n\,.\,\sum\limits_{r = 1}^n {(r - 1)\,.\,{}^{n - 1}{C_{r - 1}} + n\,.\,\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} }$

$= n\,.\,(n - 1)\,.\,{2^{n - 2}} + n\,.\,{2^{n - 1}}$]

$= 2\left( {n(n - 1){2^{n - 2}} + n\,.\,{2^{n - 1}}} \right) - n\,.\,{2^{n - 1}}$

Put $n = 10$

$= 2\left( {10\,.\,9\,.\,{2^8} + 10\,.\,{2^9}} \right) - 10\,.\,{2^9}$

$= 45\,.\,{2^{10}} + 10\,.\,{2^{10}} - 5\,.\,{2^{10}}$

$= {2^{10}}(45 + 10 - 5)$

$= {2^{10}}\,.\,(50)$

$= 25\,.\,{2^{11}}$

R.H.S. :-

${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} +$ ..... upto 10 terms)

${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${{{C_0}} \over 1} + {{{C_1}} \over 2} + {{{C_2}} \over 3} +$ ..... upto 10 terms)

${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^n{C_r}} \over {r + 1}}} } \right)$

$= {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^{n + 1}{C_{r + 1}}} \over {n + 1}}} } \right)$

$= {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_1} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}}} \over {n + 1}}} \right)$

$= {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_0} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}} - {}^{n + 1}{C_0}} \over {n + 1}}} \right)$

$= {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{n + 1}} - 1} \over {n + 1}}} \right)$

Putting value of $n = 10$, we get

$= {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$

Using L.H.S. = R.H.S.

$\Rightarrow 25\,.\,{2^{11}} = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$

$\Rightarrow 25\,.\,{2^{11}} = {2^{11}}\left( {{\alpha \over {11}}} \right)\left( {{{{2^{11}} - 1} \over {{2^\beta } - 1}}} \right)$

By comparing both sides,

${\alpha \over {11}} = 25 \Rightarrow \alpha = 275$

and ${{{2^{11}} - 1} \over {{2^\beta } - 1}} = 1$

$\Rightarrow {2^{11}} = {2^\beta }$

$\Rightarrow \beta = 11$

$\therefore$ $\alpha + \beta = 275 + 11 = 286$