Question

The coefficient of $x^{100}$ in the expression of ${\displaystyle \sum _{j=0}^{200}}{(1+x)}^j$ is

• A. ${}^{200}C_{100}$
• B. ${}^{201}C_{102}$
• C. ${}^{200}C_{101}$
• D. ${}^{201}C_{100}$

Answer 1

The correct option is D

${}^{201}C_{100}$

Explanation for the correct option:

Step-1: Simplification

Given, the expression is ${\displaystyle \sum _{j=0}^{200}}{(1+x)}^j={(1+\mathrm{x})}^0+{(1+\mathrm{x})}^1+{(1+\mathrm{x})}^2+\dots +{(1+\mathrm{x})}^{200}$.

This is a geometric series with the first term $a={\left(1+x\right)}^0=1$ and the common ratio $r=\left(1+x\right)$.

Formula to be used: We know that the sum of a finite geometric series of $n$terms with the first term $a$ and the common ratio $r$ is $S_n=\frac{a\left(r^n-1\right)}{r-1}$.

Here, $a=1$, $r=\left(1+x\right)$and $n=201$. Therefore, we get:

$$\begin{gathered} {\displaystyle \sum _{j=0}^{200}}{(1+x)}^j={(1+\mathrm{x})}^0+{(1+\mathrm{x})}^1+{(1+\mathrm{x})}^2+\dots +{(1+\mathrm{x})}^{200} \\ ={\displaystyle \frac{1\left[{\left(1+x\right)}^{201}-1\right]}{1+x-1}} \\ ={\displaystyle \frac{{\left(1+x\right)}^{201}-1}{x}} \end{gathered}$$

Step-2: Finding the coefficients of $x^{100}$

Therefore, we have to find the coefficient of ${\mathrm{x}}^{100}$ in $\frac{{(1+\mathrm{x})}^{201}-1}{\mathrm{x}}$.

Now, the coefficients of ${\mathrm{x}}^{100}$ in $\frac{{(1+\mathrm{x})}^{201}-1}{\mathrm{x}}$ is equal to the coefficient of ${\mathrm{x}}^{101}$ in ${\left(1+x\right)}^{201}$.

Formula to be used: We know that the coefficient of ${\mathrm{x}}^{\mathrm{r}}$ in the binomial expansion of ${\left(1+x\right)}^n$ is ${}^{n}C_r$.

So, the coefficient of ${\mathrm{x}}^{101}$ in ${\left(1+x\right)}^{201}$ is ${}^{201}\mathrm{C}_{101}={}^{201}\mathrm{C}_{100}\left[\because {}^{n}C_r={}^{n}C_{n-r}\right]$.

Hence, option (D) is correct answer.