The sum of the coefficients of x^499 and x^500 in (1 + x)^1000 + x(1 + x)^999 + x^2(1 + x)…

Question

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \ldots + x^{1000}$ is :

Accepted Answer

Let

$S=(1+x{)}^{1000}+x(1+x{)}^{999}+{x}^{2}(1+x{)}^{998}+\cdots +{x}^{1000}$

Notice that this is a Geometric Progression (GP) with:

First term (a): $(1+x{)}^{1000}$

Common ratio $(r):\frac{x}{1+x}$

Number of terms ( $n$ ): 1001

The sum of a GP is given by $S=a\frac{(1-{r}^{n})}{(1-r)}$. Substituting our values:

$S=(1+x{)}^{1000}[\frac{1-{(\frac{x}{1+x})}^{1001}}{1-\frac{x}{1+x}}]$

Simplify the denominator: $1-\frac{x}{1+x}=\frac{1+x-x}{1+x}=\frac{1}{1+x}$.

$S=(1+x{)}^{1000}\cdot (1+x)[1-\frac{{x}^{1001}}{(1+x{)}^{1001}}]$

$S=(1+x{)}^{1001}-{x}^{1001}$

Now we need the coefficients of ${x}^{499}$ and ${x}^{500}$ in the simplified expression $(1+x{)}^{1001}-{x}^{1001}$.

Since ${x}^{1001}$ only affects the term with power 1001, we only look at $(1+x{)}^{1001}$.

In the binomial expansion $(1+x{)}^{n}$

Coefficient of ${x}^{r}$ is ${}^{n}{C}_{r}$

Coefficient of ${x}^{499}$ in $S$ :

${}^{1001}{C}_{499}$

Coefficient of ${x}^{500}$ in $S$ :

${}^{1001}{C}_{500}$ (The term ${x}^{1001}$ does not contribute.)

$\begin{gathered} {}^{1001}{C}_{499}+{}^{1001}{C}_{500}={}^{1002}{C}_{500} \end{gathered}$

[

${}^{n}{C}_{r}+{}^{n}{C}_{r+1}={}^{n+1}{C}_{r+1}$]

Solutions · 1