First slide

Binomial theorem for positive integral Index

Question

If sum of the coefficients in the expansion of ${(x + \frac{1}{x})}^{n}$ is 128., then coefficient of x in the expansion of ${(x + \frac{1}{x})}^{n}$ is

Moderate

A

51
B

48
C

35
D

28

Solution

Sum of the coefficients in the expansion of ${(x + \frac{1}{x})}^{n}$ is equal to $2^{n}$ .
$∴ 2^{n} = 128 = 2^{7} \Rightarrow n = 7$
Now, $T_{r + 1}$ , the ${(r + 1)}^{t h}$ term in the expansion of
${(x + \frac{1}{x})}^{7}$ is
$T_{r + 1} =^{7} C_{r} x^{7 - r} {(\frac{1}{x})}^{r} =^{7} C_{r} x^{7 - 2 r}$
For the coefficient of x, we set $7 - 2 r = 1 \Rightarrow r = 3 .$
Thus, coefficient of x in the expansion of ${(x + \frac{1}{x})}^{7}$ is $^{7} C_{3} = 35$

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