Let C_r denote the binomial coefficient of x^r in the expansion of ${(1 + x)^{10}}$. If for $\alpha$, $\beta$ $\in$ R, ${C_1} + 3.2{C_2} + 5.3{C_3} +$ ....... upto 10 terms $$ = {{\alpha \times {2^{11}}} \over {{2^\beta } - 1}}\left( {{C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + \,\,.....\,\,upto\,10\,terms} \right)$$ then the value of $\alpha$ + $\beta$ is equal to ___________.

<p>Given,</p> <p>${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} +$ ...... upto 10 terms</p> <p>$= {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3}$ + ..... upto 10 terms)</p> <p>Now,</p> <p>L.H.S. :-</p> <p>${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} +$ ...... upto 10 terms</p> <p>$= 1\,.\,1{C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} +$ ..... upto 10 terms</p> <p>$= \sum\limits_{r = 1}^{10} {r\,.\,(2r - 1){}^{10}{C_r}}$</p> <p>$= \sum\limits_{r = 1}^{10} {(2{r^2} - r)\,.\,{}^{10}{C_r}}$</p> <p>$$ = 2\,.\,\sum\limits_{r = 1}^{10} {{r^2}\,.\,{}^{10}{C_r} - \sum\limits_{r = 1}^{10} {r\,.\,{}^n{C_r}} } $$</p> <p>[We know, $\sum\limits_{r = 1}^n {r\,.\,{}^n{C_r} = n\,.\,{2^{n - 1}}}$</p> <p>and $$\sum\limits_{r = 1}^n {{r^2}\,.\,{}^n{C_r} = \sum\limits_{r = 1}^n {(r\,.\,{}^n{C_r})\,.\,r} } $$</p> <p>$$ = \sum\limits_{r = 1}^n {\left( {r\,.\,{n \over r}\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r} $$</p> <p>$= \sum\limits_{r = 1}^n {\left( {n\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r}$</p> <p>$= n\sum\limits_{r = 1}^n {(r - 1 + 1){}^{n - 1}{C_{r - 1}}}$</p> <p>$$ = n\,.\,\sum\limits_{r = 1}^n {(r - 1)\,.\,{}^{n - 1}{C_{r - 1}} + n\,.\,\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} } $$</p> <p>$= n\,.\,(n - 1)\,.\,{2^{n - 2}} + n\,.\,{2^{n - 1}}$]</p> <p>$= 2\left( {n(n - 1){2^{n - 2}} + n\,.\,{2^{n - 1}}} \right) - n\,.\,{2^{n - 1}}$</p> <p>Put $n = 10$</p> <p>$= 2\left( {10\,.\,9\,.\,{2^8} + 10\,.\,{2^9}} \right) - 10\,.\,{2^9}$</p> <p>$= 45\,.\,{2^{10}} + 10\,.\,{2^{10}} - 5\,.\,{2^{10}}$</p> <p>$= {2^{10}}(45 + 10 - 5)$</p> <p>$= {2^{10}}\,.\,(50)$</p> <p>$= 25\,.\,{2^{11}}$</p> <p>R.H.S. :-</p> <p>${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} +$ ..... upto 10 terms)</p> <p>${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${{{C_0}} \over 1} + {{{C_1}} \over 2} + {{{C_2}} \over 3} +$ ..... upto 10 terms)</p> <p>$${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^n{C_r}} \over {r + 1}}} } \right)$$</p> <p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^{n + 1}{C_{r + 1}}} \over {n + 1}}} } \right)$$</p> <p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_1} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}}} \over {n + 1}}} \right)$$</p> <p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_0} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}} - {}^{n + 1}{C_0}} \over {n + 1}}} \right)$$</p> <p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{n + 1}} - 1} \over {n + 1}}} \right)$$</p> <p>Putting value of $n = 10$, we get</p> <p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$$</p> <p>Using L.H.S. = R.H.S.</p> <p>$$ \Rightarrow 25\,.\,{2^{11}} = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$$</p> <p>$$ \Rightarrow 25\,.\,{2^{11}} = {2^{11}}\left( {{\alpha \over {11}}} \right)\left( {{{{2^{11}} - 1} \over {{2^\beta } - 1}}} \right)$$</p> <p>By comparing both sides,</p> <p>${\alpha \over {11}} = 25 \Rightarrow \alpha = 275$</p> <p>and ${{{2^{11}} - 1} \over {{2^\beta } - 1}} = 1$</p> <p>$\Rightarrow {2^{11}} = {2^\beta }$</p> <p>$\Rightarrow \beta = 11$</p> <p>$\therefore$ $\alpha + \beta = 275 + 11 = 286$</p>