First slide

Binomial theorem for positive integral Index

Question

The number of irrational terms in the expansion of ${(5^{1 / 6} + 2^{1 / 8})}^{100}$ is

Moderate

A

$96$
B

$97$
C

$98$
D

$99$

Solution

$T_{r + 1},$ the $(r + 1)$ th the expansion of ${(5^{1 / 6} + 2^{1 / 8})}^{100}$ is given by
$T_{r + 1} =^{100} C_{r} {(5^{1 / 6})}^{100 - r} {(2^{1 / 8})}^{r}$
As $5$ and $2$ are relatively prime, $T_{r + 1}$ will be rational if $\frac{100 - r}{6}$ and $\frac{r}{8}$ are both integers. i.e. if $100 - r$ is a multiple of $6$ and $r$ is multiple of $8 .$ As $0 \leq r \leq 100,$ multiples of $8$ upto $100$ and corresponding value of $100 - r$ are
$\begin{aligned} r = 0, 8, 16, 24, \dots, 88, 96 \\ 100 - r = 100, 92, 84, 76, \dots, 12, 4 \end{aligned}$
Out of these values for $100 - r$ , multiples of $6$ are $84, 60, 36, 12$
$∴$ there are just four rational terms in the expansion of ${(5^{1 / 6} + 2^{1 / 8})}^{100}$ .
$\Rightarrow$ number of irrational terms is $101 - 4 = 97$

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