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

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

$$ \Rightarrow \beta = 11$$

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