First slide

Binomial theorem for positive integral Index

Question

The ratio of coefficient of $x^{15}$ to the constant term in the expansion of ${(x^{2} + \frac{2}{x})}^{15}$ is

Easy

A

$\frac{1}{16}$
B

$\frac{1}{32}$
C

$\frac{1}{4}$
D

None of these

Solution

Given expansion is   ${(x^{2} + \frac{2}{x})}^{15}$
We have general term in the expansion ${(x + a)}^{n}$

$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$

General term   $T_{r + 1} =^{15} C_{r} {(x^{2})}^{15 - r} {(\frac{2}{x})}^{r}$

$T_{r + 1} =^{15} C_{r} {(x)}^{30 - 2 r} {(2)}^{r} {(x^{- 1})}^{r}$

        $T_{r + 1} =^{15} C_{r} x^{30 - 3 r} 2^{r}$

Compare $x^{30 - 3 r}$ with $x^{15}$

$\Rightarrow x^{30 - 3 r} = x^{15}$

This contains $x^{15}$ if $30 - 3 r = 15$ i.e. if $r = 5$

$∴$ Coefficient of $x^{15}$ $=^{15} C_{5} 2^{5} = α$ (say)

Also, $T_{r + 1}$ is void of $x$ if $\Rightarrow x^{30 - 3 r} = x^{0}$

( the constant term in the expansion compare with $x^{0}$ )

$30 - 3 r = 0$ i.e. if $r = 10$

$∴$ Constant term $=^{15} C_{10} 2^{10} = b$ (say)

Hence, the required ratio $= \frac{a}{b} = \frac{^{15} C_{5} 2^{5}}{^{15} C_{10} 2^{10}}$

                                                $= \frac{1}{2^{5}} = \frac{1}{32}$ .                  $(∵^{15} C_{5} =^{15} C_{10})$

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