Jee Main 2024 30 01 2024 Shift1 - Question15
Question 15
Number of integral terms in the binomial expansion of $\left(7^{1 / 2}+11^{1 / 6}\right)^{824}$ is
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Answer: (138)
Solution:
$T _{n+1}={ }^{n} C _r 11^{\frac{r}{6}} \cdot 7^{\frac{824-r}{2}}$
For integral term
6 should divide $r$
and $\frac{824-r}{2}$ must be integer
$\Rightarrow 2$ most divide $r$
$\Rightarrow r$ divisible by 6
$\Rightarrow$ possible values of $r \in{0,1,2, \ldots 824}$
$\Rightarrow$ For integer terms
$r \in{0,6,12, \ldots 822}(822=0+(n-1) 6 \Rightarrow n=138)$
$=138$ terms
BETA
