Let the smallest value of \(\mathrm{k} \in \mathbb{N}\), for which the coefficient of \(\mathrm{x}^{3}\) in \((1+\mathrm{x})^{3}+(1+\mathrm{x})^{4}+(1+\mathrm{x})^{5}+\ldots +(1+x)^{99}+(1+k x)^{100}, x \neq 0\), is \(\left(43 n+\frac{101}{4}\right)\) ( \({ }^{100} \mathrm{C}_{3}\) ) for some \(\mathrm{n} \in \mathbb{N}\), be \(p\). Then the value of \(p\) +n is:

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Binomial Theorem

Let the smallest value of

k \in N

, for which the coefficient of

x^{3}

(1 + x)^{3} + (1 + x)^{4} + (1 + x)^{5} + \dots + (1 + x)^{99} + (1 + k x)^{100}, x \neq = 0

, is

(43 n + \frac{101}{4})

(

^{100} C_{3}

) for some

n \in N

, be

p

. Then the value of

p

+n is:

Answer: (B)

The coefficient of

x^{3}

m = 3 \sum 99 (3 m) + k^{3} (3 100)

, and by the hockey-stick identity

m = 3 \sum 99 (3 m) = (4 100)

. Dividing by

(3 100)

and using

\frac{( 4 100 )}{( 3 100 )} = \frac{97}{4}

gives

\frac{97}{4} + k^{3} = 43 n + \frac{101}{4}

, so

k^{3} = 43 n + 1

. The least

k \in N

making

43 n + 1

a perfect cube with

n \in N

k = 6

, since

6^{3} = 216 = 43 (5) + 1

, whence

n = 5

. Therefore

p = 6

n = 5

and

p + n = 11

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option