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7.Binomial Theorem
hard
The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is equal to :
A
$3133$
B
$633$
C
$931$
D
$6131$
(JEE MAIN-2024)
Solution
$ \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(5^{\frac{1}{3}}\right)^{\mathrm{r}}\left(2^{\frac{1}{5}}\right)^{15-\mathrm{r}} $
$ ={ }^{15} \mathrm{C}_{\mathrm{r}} 5^{\frac{\mathrm{r}}{3}} \cdot 2^{\frac{15-\mathrm{r}}{5}} $
$ \mathrm{R}=3 \lambda, 15 \mu $
$ \Rightarrow \mathrm{r}=0,15 $
$ 2 \text { rational terms } $
$ \Rightarrow{ }^{15} \mathrm{C}_0 2^3+{ }^{15} \mathrm{C}_{15}(5)^5 $
$ =8+3125=3133$
Standard 11
Mathematics