If the constant term in the expansion of ((√53)/()x^+)12,x 0, is 2^8 √53, then 25 is equal…

Question

If the constant term in the expansion of ${(\frac{\sqrt53}{}x}^{+})12,x
e 0$, is $\alpha 	imes {2}^{8}	imes \sqrt53$, then $25\alpha $ is equal to :

Accepted Answer

$\begin{gathered} {(\frac{\sqrt53}{}x}^{+})12 \end{gathered}{T}_{r+1}={}^{12}Cr{(\frac{\sqrt53}{}x}^{)}12-r{(\frac{2x}{\sqrt35}}^{)}r$

For constant term $-12+r+r=0$

$\begin{gathered} \Rightarrow \, r=6 \ 	herefore \, ={}^{12}{C}_{6}\frac{(3{)}^{\frac{6}{5}}}{(5{)}^{\frac{6}{3}}}(2{)}^{6} \end{gathered}$

$\begin{gathered} ={}^{12}{C}_{6}	imes \frac{{2}^{6}}{25}	imes {3.3}^{\frac{1}{5}} \ =\frac{231}{25}	imes {2}^{8}\cdot {3}^{\frac{1}{5}}\cdot 3 \ =\frac{693}{25}\cdot {2}^{8}\sqrt53 \end{gathered}	herefore \, \alpha =\frac{693}{25}25\alpha =693$

Solutions · 1