The ratio of the coefficient of x

Question:

The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in $\left(x^{2}+\frac{2}{x}\right)^{15}$, is

(a) 12 : 32

(b) 1 : 32

(c) 32 : 12

(d) 32 : 1

Solution:

$\ln \left(x^{2}+\frac{2}{x}\right)^{15}$ we have $T_{r+1}={ }^{15} C_{r}\left(x^{2}\right)^{15-r}\left(\frac{2}{x}\right)^{r}$

i.e. General term is $T_{r+1}={ }^{15} C_{r} x^{30-3 r} 2^{r}$

Hence for the term independent of x,

30 – 3= 0

i.e. r = 10

hence T11 has coefficient 15C10 210       ...(1)

and term with x15 will have 30 – 3= 15

i.e. 15 = 3r

i.e. r = 5

∴ coefficient will be 15C25              ...(2)

∴  ratio of coefficient of x15 to the term independent of x will be

$\frac{e q^{n}(2)}{e q^{n}(1)}=\frac{2^{5} \times{ }^{15} C_{5}}{2^{10} \times{ }^{15} C_{10}}$

$=\frac{1}{2^{5}} \quad\left(\because{ }^{15} C_{10}={ }^{15} C_{5-10}={ }^{15} C_{5}\right)$

i.e. ratio will be 1 : 32

Hence, the correct answer is option B.

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