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7.Binomial Theorem
normal
The ratio of coefficient of $x^2$ to coefficient of $x^{10}$ in the expansion of ${\left( {{x^5} + {{4.3}^{ - {{\log }_{\sqrt 3 }}\sqrt {{x^3}} }}} \right)^{10}}$ is
A
$4:7$
B
$10:3$
C
$3:10$
D
$7:4$
Solution
${\left( {{x^5} + {{4.3}^{ – {{\log }_{\sqrt 3 }}\sqrt {{x^3}} }}} \right)^{10}} = {\left( {{x^5} + 4 \cdot {x^{ – 3}}} \right)^{10}}$
${{\rm{T}}_{{\rm{r}} + 1}} = {\,^{10}}{{\rm{C}}_{\rm{r}}}{\left( {{{\rm{x}}^5}} \right)^{10 – {\rm{r}}}}{\left( {4{{\rm{x}}^{ – 3}}} \right)^r}$
$ = {\,^{10}}{{\rm{C}}_{\rm{r}}}{4^r}{{\rm{x}}^{50 – 5{\rm{r}} – 3{\rm{r}}}}$
$ = {\,^{10}}{{\rm{C}}_r}{{\rm{4}}^r}{{\rm{x}}^{50 – 8{\rm{r}}}}$
$50-8 r=2 \Rightarrow r=6$
and $50-8 r=10 \Rightarrow r=5$
Required ratio
$ = \frac{{^{10}{C_6}{4^6}}}{{^{10}{C_5}{4^5}}} = \frac{{10}}{3}$
Standard 11
Mathematics
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