The coefficient of x301 in the expansion of (1+x)500+x(1+x)499+x2(1+x)498+ ldots . .+x500 is

Question

The coefficient of x301 in the expansion of (1+x)500+x(1+x)499+x2(1+x)498+ ldots . .+x500 is (A)  501 C301 (B)  500 C301 (C)  501 C300 (D) none of these

Tardigrade Admin · Accepted Answer

The given series S =(1+x)500[1+(x/1+x)+((x/1+x))2+ ldots .+((x/1+x))500] =(1+x)500 × (1-((x/1+x))501/1-(x)1+x)=(1+x)501-x501 Hence, the coefficient of x301 in S=501 C 301 .

Q. The coefficient of $x^{301}$ in the expansion of $(1 + x)^{500} + x (1 + x)^{499} + x^{2} (1 + x)^{498} + \dots .. + x^{500}$ is

$^{501} C_{301}$

$^{500} C_{301}$

$^{501} C_{300}$

none of these

The given series
$S = (1 + x)^{500} [1 + \frac{x}{1 + x} + (\frac{x}{1 + x})^{2} + \dots . + (\frac{x}{1 + x})^{500}]$
$= (1 + x)^{500} \times \frac{1 - ( \frac{x}{1 + x} ) ^{501}}{1 - \frac{x}{1 + x}} = (1 + x)^{501} - x^{501}$
Hence, the coefficient of $x^{301}$ in $S = 501 C 301.$

Q. The coefficient of x301 in the expansion of (1+x)500+x(1+x)499+x2(1+x)498+…..+x500 is

501C301​

500C301​

501C300​