If 26(2^33(^12C_2)+2^55(^12C_4)+2^77(^12C_6)+ +2^1313(^12C_12))=3^13- , then is equal to :

Question

If $26(\frac{{2}^{3}}{3}({}^{12}{C}_{2})+\frac{{2}^{5}}{5}({}^{12}{C}_{4})+\frac{{2}^{7}}{7}({}^{12}{C}_{6})+\cdots +\frac{{2}^{13}}{13}({}^{12}{C}_{12}))={3}^{13}-\alpha $, then $\alpha $ is equal to :

Accepted Answer

${S}_{r}=26[(\frac{{2}^{3}}{3})\cdot ({}^{12}{C}_{2})+(\frac{{2}^{5}}{5})\cdot ({}^{12}{C}_{4})+(\frac{{2}^{7}}{7})\cdot ({}^{12}{C}_{6})+(\frac{{2}^{9}}{9})\cdot ({}^{12}{C}_{8})+\ldots .+(\frac{{2}^{13}}{13})\cdot ({}^{12}{C}_{12})]$

$=26[\sum _{r=1}^{6}(\frac{{2}^{2r+1}}{2r+1})\cdot ({}^{12}{C}_{2r})]$.

${S}_{r}=\frac{26}{13}[\sum _{r=1}^{6}({2}^{2r+1})\cdot ({}^{13}{C}_{2r+1})]$

$=2[\sum _{r=0}^{6}({2}^{2r+1})\cdot ({}^{13}{C}_{2r+1})-({}^{13}{C}_{1})\cdot {2}^{1}]$

$=2[\sum _{r=0}^{6}({2}^{2r+1})\cdot ({}^{13}{C}_{2r+1})]-52$

$\begin{gathered} {S}_{r} =2[({}^{13}{C}_{1})\cdot {2}^{1}+({}^{13}{C}_{3})\cdot {2}^{3}+({}^{13}{C}_{5})\cdot {2}^{5}+({}^{13}{C}_{7})\cdot {2}^{7}+\ldots ..+({}^{13}{C}_{13})\cdot {2}^{13}]-52 \ =(1+2{)}^{13}-(1-2{)}^{13}-52 \ ={3}^{13}+1-52 \ ={3}^{13}-51.....(i) \end{gathered}$

${S}_{r}=26[(\frac{{2}^{3}}{3})\cdot ({}^{12}{C}_{2})+(\frac{{2}^{5}}{5})\cdot ({}^{12}{C}_{4})+(\frac{{2}^{7}}{7})\cdot ({}^{12}{C}_{6})+(\frac{{2}^{9}}{9})\cdot ({}^{12}{C}_{8})+\ldots .+(\frac{{2}^{13}}{13})\cdot ({}^{12}{C}_{12})]={3}^{13}-\alpha \ldots .(ii)$

From the equivalence relationship between equation $(i)$ and $(ii)$ we conclude $\alpha =51$

Solutions · 1