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Q. The sum of the coefficients in the expansion of $\left(x^{2}-\frac{1}{3}\right)^{199} \times\left(x^{3}+\frac{1}{2}\right)^{200}$ is

KEAMKEAM 2011Binomial Theorem

Solution:

Given expression $\left(x^{2}-\frac{1}{3}\right)^{199} \times\left(x^{3}+\frac{1}{2}\right)^{200}$
The sum of the coefficients in the above expression
$=\left(1-\frac{1}{3}\right)^{199} \times\left(1+\frac{1}{2}\right)^{200}$
$(\because$ put $x=1)$
$=\left(\frac{2}{3}\right)^{199} \times\left(\frac{3}{2}\right)^{200} $
$=\left(\frac{3}{2}\right)^{(200-199)}=\frac{3}{2}$