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7.Binomial Theorem
hard
The sum of the rational terms in the binomial expansion of ${\left( {{2^{\frac{1}{2}}} + {3^{\frac{1}{5}}}} \right)^{10}}$ is
A
$25$
B
$32$
C
$9$
D
$41$
(JEE MAIN-2013)
Solution
$\left(2^{1 / 2}+3^{1 / 5}\right)^{10}=^{10} \mathrm{C}_{0}\left(2^{1 / 2}\right)^{10}$
$+^{10} \mathrm{C}_{1}\left(2^{1 / 2}\right)^{9}\left(3^{1 / 5}\right)+\ldots \ldots+^{10} \mathrm{C}_{10}\left(3^{1 / 5}\right)^{10}$
There are onlytwo rational terms – first term and last term.
Now sum of two rational terms
$=(2)^{5}+(3)^{2}=32+9=41$
Standard 11
Mathematics
