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) + ... + (1+x^99) + (1+kx)^100, x != 0, is (43n + 101/4) C(100,3) for some n in N, be p. Then the value of p+n is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q9 – JEE-MAIN 2026

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Home Maths Binomial TheoremQuestion #37

Let the smallest value of

k \in N

, for which the coefficient of

x^{3}

(1 + x^{3}) + (1 + x^{4}) + (1 + x^{5}) + . . . + (1 + x^{99}) + (1 + k x)^{100}

x \neq 0

, is

(43 n + \frac{101}{4}) (\binom{100}{3})

for some

n \in N

, be

p

. Then the value of

p + n

is:

Clue 1 of 3: Binomial Expansion and Series Summation

The problem involves finding the coefficient of a specific term in a sum of binomial expansions. Note that there is a likely typo in the question, and the terms $(1 + x^{m})$ should be interpreted as $(1 + x)^{m}$ for the sum part to yield an integer solution.

🧠 Full Solution Path

Step 1: Identify the coefficient of

x^{3}

in the given expression✦ Active

The given expression is a sum of terms. Assuming a common typo, the terms $(1 + x^{m})$ in the series should be interpreted as $(1 + x)^{m}$ for $m = 3, 4, \dots, 99$ . Otherwise, an integer solution for $k$ is not possible. The last term is $(1 + k x)^{100}$ . We need to find the coefficient of $x^{3}$ in this entire sum.

For the sum $\sum_{m = 3}^{99} (1 + x)^{m}$ , the coefficient of $x^{3}$ in each term $(1 + x)^{m}$ is $(\binom{m}{3})$ (for $m \geq 3$ ). The sum of these coefficients is:

\sum_{m = 3}^{99} (\binom{m}{3}) = (\binom{3}{3}) + (\binom{4}{3}) + \dots + (\binom{99}{3})

Using the Hockey-stick identity $\sum_{i = r}^{N} (\binom{i}{r}) = (\binom{N + 1}{r + 1})$ , with $r = 3$ and $N = 99$ :

\sum_{m = 3}^{99} (\binom{m}{3}) = (\binom{99 + 1}{3 + 1}) = (\binom{100}{4})

For the term $(1 + k x)^{100}$ , the coefficient of $x^{3}$ is $(\binom{100}{3}) (k)^{3}$ . Therefore, the total coefficient of $x^{3}$ in the entire expression is:

C_{x^{3}} = (\binom{100}{4}) + (\binom{100}{3}) k^{3}

💡 Teacher's Secret Hint

Carefully distinguish between $(1 + x^{m})$ and $(1 + x)^{m}$ . The problem's structure strongly suggests the latter for a solvable integer outcome.

Step 2: Formulate and simplify the equation for

k

and

n

○ Expand

We are given that the coefficient of $x^{3}$ is $(43 n + \frac{101}{4}) (\binom{100}{3})$ . Equating this with our derived coefficient:

(\binom{100}{4}) + (\binom{100}{3}) k^{3} = (43 n + \frac{101}{4}) (\binom{100}{3})

Divide the entire equation by $(\binom{100}{3})$ :

\frac{(\binom{100}{4})}{(\binom{100}{3})} + k^{3} = 43 n + \frac{101}{4}

Using the identity $\frac{(\binom{N}{R})}{(\binom{N}{R - 1})} = \frac{N - R + 1}{R}$ , for $N = 100$ and $R = 4$ :

\frac{(\binom{100}{4})}{(\binom{100}{3})} = \frac{100 - 4 + 1}{4} = \frac{97}{4}

Substitute this back into the equation:

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

Rearrange to solve for $k^{3}$ :

k^{3} = 43 n + \frac{101}{4} - \frac{97}{4} = 43 n + \frac{4}{4} = 43 n + 1

💡 Teacher's Secret Hint

Remember the binomial coefficient identities to simplify the ratio of combinations efficiently.

Step 3: Find the smallest

k

and corresponding

n

, then calculate

p + n

○ Expand

We need to find the smallest value of $k \in N$ (natural numbers, $k \geq 1$ ) such that $k^{3} = 43 n + 1$ for some $n \in N$ (natural numbers, $n \geq 1$ ). Let's test values for $n$ :

- For $n = 1$ , $k^{3} = 43 (1) + 1 = 44$ (not a perfect cube).

- For $n = 2$ , $k^{3} = 43 (2) + 1 = 86 + 1 = 87$ (not a perfect cube).

- For $n = 3$ , $k^{3} = 43 (3) + 1 = 129 + 1 = 130$ (not a perfect cube).

- For $n = 4$ , $k^{3} = 43 (4) + 1 = 172 + 1 = 173$ (not a perfect cube).

- For $n = 5$ , $k^{3} = 43 (5) + 1 = 215 + 1 = 216$ . Since $216 = 6^{3}$ , this is a perfect cube.

The smallest value of $k$ is $6$ , which occurs when $n = 5$ . Therefore, $p = 6$ and $n = 5$ . The question asks for the value of $p + n$ .

p + n = 6 + 5 = 11

💡 Teacher's Secret Hint

Systematically check values of $n$ starting from $1$ to find the first instance where $43 n + 1$ is a perfect cube. This will give the smallest $k$ .

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Maths Question 9 – JEE-MAIN 2026

Clue 1 of 3: Binomial Expansion and Series Summation

🧠 Full Solution Path