JEE Main 2026 — Apr 4, Shift 1 — Mathematics solution

JEE Main2026Apr 4, Shift 1Mathematics

Let the smallest value of $k \in N$ , for which the coefficient of $x^{3}$ in $(1 + x)^{3} + (1 + x)^{4} + (1 + x)^{5} + \dots + (1 + x)^{99} + (1 + k x)^{100}$ , $x \neq = 0$ , is $(43 n + \frac{101}{4}) (3 100)$ for some $n \in N$ , be $p$ . Then the value of $p + n$ is:

A
$10$
B
$11$
C
$12$
D
$13$

Solution

The $x^{3}$ coefficient from $(1 + x)^{3} + \dots + (1 + x)^{99}$ is $\sum_{r = 3}^{99} (3 r)$ . Using the hockey-stick identity,

$\sum_{r = 3}^{99} (3 r) = (4 100) .$

From $(1 + k x)^{100}$ , the $x^{3}$ coefficient is $(3 100) k^{3}$ . So the total coefficient is

$(4 100) + (3 100) k^{3} = (\frac{( 4 100 )}{( 3 100 )} + k^{3}) (3 100) .$

Since $\frac{( 4 100 )}{( 3 100 )} = \frac{97}{4}$ , the bracket equals $\frac{97}{4} + k^{3}$ . Setting this equal to $43 n + \frac{101}{4}$ gives

$k^{3} = 43 n + 1.$

Search for the smallest $k \in N$ making $k^{3} - 1$ divisible by $43$ . Try $k = 6$ : $k^{3} - 1 = 215 = 43 \cdot 5$ , so $n = 5$ .

$∴ p = 6, n = 5, p + n = 11.$

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