Q. The sum of the coefficients of $x^{499}$ and $x^{500}$ in

$(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}$

is :

Correct Answer: ${1002 \choose 500}$

Explanation

Coefficient of $x^r$ in $(1+x)^n$ is ${n \choose r}$.

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From $(1+x)^{1000}$ :

Coefficient of $x^{499} = {1000 \choose 499}$

Coefficient of $x^{500} = {1000 \choose 500}$

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From $x(1+x)^{999}$ :

Coefficient of $x^{499} = {999 \choose 498}$

Coefficient of $x^{500} = {999 \choose 499}$

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From $x^{2}(1+x)^{998}$ :

Coefficient of $x^{499} = {998 \choose 497}$

Coefficient of $x^{500} = {998 \choose 498}$

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Now sum all coefficients:

$[{1000 \choose 499}+{999 \choose 498}+{998 \choose 497}] + [{1000 \choose 500}+{999 \choose 499}+{998 \choose 498}]$

Using Pascal’s identity repeatedly,

$= {1002 \choose 500}$

Hence, the required sum is

${1002 \choose 500}$

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