First slide

Binomial theorem for positive integral Index

Question

The ratio of the coefficient of $x^{10}$ in ${(1 + x^{2})}^{10}$ and the term independent of $x$ in ${(x + \frac{2}{x})}^{10}$ is

Moderate

A

32 : 1
B

1 : 32
C

1 : 64
D

64 : 1

Solution

$T_{r + 1} = (r + 1)$ th term in ${(1 + x^{2})}^{10} =^{10} C_{r} {(x^{2})}^{r}$
$∴ coefficient of x^{10} in {(1 + x^{2})}^{10} {is}^{10} C_{5} .$
Also, $t_{r + 1} = (r + 1)$ th term in ${(x + \frac{2}{x})}^{10}$
$\begin{array}{l} =^{10} C_{r} (x)^{10 - r} {(\frac{2}{x})}^{r} \\ =^{10} C_{r} 2^{r} x^{10 - 2 r} \end{array}$
For this term to be independent of $x,$ set $10 - 2 r = 0$
or $r = 5 .$
$∴$ term independent of $x$ is $^{10} C_{5} (2^{5})$

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