The coefficient of x^{301} in (1+x)^{500} + x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given expression is a geometric series with first term $a = (1+x)^{500}$,common ratio $r = \frac{x}{1+x}$,and number of terms $n = 501$.The su

Vedclass · Accepted Answer

$^{501}C_{200}$

Vedclass · Answer

(D) The given expression is a geometric series with first term $a = (1+x)^{500}$,common ratio $r = \frac{x}{1+x}$,and number of terms $n = 501$.The sum of the series is $S = a \frac{1-r^n}{1-r} = (1+x)^{500} \left[ \frac{1 - (\frac{x}{1+x})^{501}}{1 - \frac{x}{1+x}} \right]$.Simplifying the expression:$S = (1+x)^{500} \left[ \frac{\frac{(1+x)^{501} - x^{501}}{(1+x)^{501}}}{\frac{1+x-x}{1+x}} \right] = (1+x)^{500} \cdot \frac{(1+x)^{501} - x^{501}}{(1+x)^{501}} \cdot (1+x) = (1+x)^{501} - x^{501}$.We need the coefficient of $x^{301}$ in $(1+x)^{501} - x^{501}$.The coefficient of $x^{301}$ in $(1+x)^{501}$ is $^{501}C_{301}$.Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we have $^{501}C_{301} = ^{501}C_{501-301} = ^{501}C_{200}$.

The coefficient of $x^{301}$ in $(1+x)^{500} + x(1+x)^{499} + x^2(1+x)^{498} + \ldots + x^{500}$ is:

Similar Questions

If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$ when $x=\frac{6}{371}$, is

Find the remainder when $3 \times 7^{22} + 2 \times 10^{22} - 44$ is divided by $18$.

Using the binomial theorem, the value of $(0.999)^3$ correct to $3$ decimal places is:

The sum of the coefficients of the terms of degree $m$ in the expansion of $(1 + x)^n(1 + y)^n(1 + z)^n$ is:

If $a_k$ is the coefficient of $x^k$ in the expansion of $(1+x+x^2)^n$ for $k=0, 1, 2, \ldots, 2n$,then $a_1+2a_2+3a_3+\ldots+2na_{2n}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module