(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+ldots | vedclass

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Question

$(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots .+x^{500}$

$=(1+x)^{500} \cdot\left\{\frac{1-\left(\frac{x}{1+x}\right)^{501}}{1-\frac{x}{1+x}}\rig

Vedclass · Accepted Answer

${ }^{501} C _{200}$

Vedclass · Answer

$(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots .+x^{500}$

$=(1+x)^{500} \cdot\left\{\frac{1-\left(\frac{x}{1+x}\right)^{501}}{1-\frac{x}{1+x}}\right\}$

$=(1+x)^{500} \frac{\left((1+x)^{501}-x^{501}\right)}{(1+x)^{501}} \cdot(1+x)$

$=(1+x)^{501}-x^{501}$

Coefficient of $x^{301}$ in $(1+x)^{501}-x^{501}$ is given by ${ }^{501} C _{301}={ }^{501} C _{200}$

$(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots . .+x^{500}$ में $\mathrm{x}^{301}$ का गुणांक है :

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