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7.Binomial Theorem
hard
${\left( {{x^2} + \frac{2}{x}} \right)^{15}}$ ના વિસ્તરણમાં $x^{15}$ ના સહગુણક અને અચળ પદનો ગુણોત્તર મેળવો.
A
$7: 16$
B
$7:64$
C
$1: 4$
D
$1: 32$
(JEE MAIN-2013)
Solution
$T_{r+1}=^{15} C_{r}\left(x^{2}\right)^{15-r} \cdot\left(2 x^{-1}\right)^r$
$=15 \mathrm{C}_{r} \times(2)^{r} \times x^{30-3r}$
For independent term, $30-3 r=0$
$\Rightarrow r=10$
Hence the term independent of $x$.
$\mathrm{T}_{11}=^{15} \mathrm{C}_{10} \times(2)^{10}$
For term involve $x^{15}, 30-3 r=15 \Rightarrow r=5$
Hence coefficient of $x^{15}=^{15} C_{5} \times(2)^{5}$
Required ratio
$ = \frac{{{\text{15}}{{\text{C}}_5} \times {{(2)}^5}}}{{15{C_{10}} \times {{(2)}^{10}}}} = \frac{{\frac{{15!}}{{10!5!}}}}{{\frac{{15!}}{{5!10!}} \times {{(2)}^5}}}$
$ = 1:32$
Standard 11
Mathematics
