Binomial Theorem Question 13
Question 13 - 30 January - Shift 1
The coefficient thon of $ x^{301}$ in $(1+x)^{500}+x(1+x)^{499}+x^{2}(1+x)^{498}+\ldots . .+x^{500}$ is:
(1) ${ }^{501} C _{302}$
(2) ${ }^{500} C _{301}$
(3) ${ }^{500} C _{300}$
(4) ${ }^{501} C _{200}$
Show Answer
Answer: (4)
Solution:
Formula: Sum of $\mathbf{n}$ terms of a G.P., Properties of Binomial Theorem for Positive Integer (The coefficient of $x^{r}$ in the expansion of $(1+x)^{n}$)
$ \begin{aligned} & (1+x)^{500}+x(1+x)^{499}+x^{2}(1+x)^{498}+\ldots .+x^{500} \\ & =(1+x)^{500} \cdot{\frac{1-(\frac{x}{1+x})^{501}}{1-\frac{x}{1+x}}} \\ & =(1+x)^{500} \frac{((1+x)^{501}-x^{501})}{(1+x)^{501}} \cdot(1+x) \\ & =(1+x)^{501}-x^{501} \end{aligned} $
Coefficient of $x^{301}$ in $(1+x)^{501}-x^{501}$ is given by
$ { }^{501} C _{301}={ }^{501} C _{200} $
