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Q.
The coefficient of $x^{301}$ in the expansion of $(1+x)^{500}+x(1+x)^{499}+x^{2}(1+x)^{498}+\ldots . .+x^{500}$ is
Binomial Theorem
Solution:
The given series
$S =(1+x)^{500}\left[1+\frac{x}{1+x}+\left(\frac{x}{1+x}\right)^{2}+\ldots .+\left(\frac{x}{1+x}\right)^{500}\right] $
$=(1+x)^{500} \times \frac{1-\left(\frac{x}{1+x}\right)^{501}}{1-\frac{x}{1+x}}=(1+x)^{501}-x^{501}$
Hence, the coefficient of $x^{301}$ in $S=501 C 301 .$