The coefficient of $x^{301}$ in $(1+x)^{500}+x(1+x)^{499}+x^{2}(1+x)^{498}+\ldots+x^{500}$ is:

(D) ${}^{501}\mathrm{C}_{200}$

(A) ${}^{501}\mathrm{C}_{302}$

(B) ${}^{500}\mathrm{C}_{301}$

(C) ${}^{500}\mathrm{C}_{300}$

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Binomial Theorem

The coefficient of

x^{301}

(1 + x)^{500} + x (1 + x)^{499} + x^{2} (1 + x)^{498} + \dots + x^{500}

is:

Answer: (D)

Each term has the form

x^{k} (1 + x)^{500 - k}

for

k = 0, 1, \dots, 500

, a geometric progression with first term

(1 + x)^{500}

and common ratio

\frac{x}{1 + x}

. Summing the

501

terms,

k = 0 \sum 500 x^{k} (1 + x)^{500 - k} = (1 + x)^{500} \cdot \frac{1 - ( \frac{x}{1 + x} ) ^{501}}{1 - \frac{x}{1 + x}} = (1 + x)^{501} - x^{501} .

Since

x^{501}

cannot contribute a power as low as

x^{301}

, the required coefficient is that of

x^{301}

(1 + x)^{501}

, namely

(301 501)

. Written with the smaller lower index, the coefficient of

x^{301}

equals

^{501} C_{200} .

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option