Smallest Value of k in Binomial Expansion Coefficient Problem

To find the smallest value of $k \in \mathbb{N}$ (denoted as $p$), we need to evaluate the coefficient of $x^3$ in the given expression: $S = (1 + x)^3 + (1 + x)^4 + (1 + x)^5 + \dots + (1 + x)^{99} + (1 + kx)^{100}$

Step 1: Find the coefficient of $x^3$

The coefficient of $x^3$ in $(1 + x)^r$ is ${}^r\text{C}_3$ for $r \ge 3$.

Summing the terms from $(1+x)^3$ to $(1+x)^{99}$: $\text{Coeff. of } x^3 \text{ in } \sum_{r=3}^{99} (1+x)^r = \sum_{r=3}^{99} {}^r\text{C}_3$

Using the binomial identity $\sum_{r=k}^n {}^r\text{C}_k = {}^{n+1}\text{C}_{k+1}$: $\sum_{r=3}^{99} {}^r\text{C}_3 = {}^{100}\text{C}_4$

The coefficient of $x^3$ in $(1 + kx)^{100}$ is: ${}^{100}\text{C}_3 \cdot k^3$

Thus, the total coefficient of $x^3$ in $S$ is: $\text{Total Coeff.} = {}^{100}\text{C}_4 + k^3 \cdot {}^{100}\text{C}_3$

Step 2: Relate ${}^{100}\text{C}_4$ and ${}^{100}\text{C}_3$

Using the combination property ${}^n\text{C}_r = \frac{n - r + 1}{r} {}^n\text{C}_{r-1}$: ${}^{100}\text{C}_4 = \frac{100 - 4 + 1}{4} {}^{100}\text{C}_3 = \frac{97}{4} {}^{100}\text{C}_3$

Substituting this back into the total coefficient expression: $\text{Total Coeff.} = \left(\frac{97}{4} + k^3\right) {}^{100}\text{C}_3$

Step 3: Equate with the given value

We are given that the coefficient of $x^3$ is $\left(43n + \frac{101}{4}\right){}^{100}\text{C}_3$: $k^3 + \frac{97}{4} = 43n + \frac{101}{4}$

Subtracting $\frac{97}{4}$ from both sides: $k^3 - 43n = \frac{101 - 97}{4} = 1$ $k^3 - 1 = 43n$

Step 4: Determine the smallest value of $k$ ($p$) and corresponding $n$

Since $n \in \mathbb{N}$, $k^3 - 1$ must be a positive multiple of $43$: $k^3 \equiv 1 \pmod{43} \quad (k > 1)$

This can be factored as: $(k - 1)(k^2 + k + 1) \equiv 0 \pmod{43}$

Since $k \not\equiv 1 \pmod{43}$, we solve: $k^2 + k + 1 \equiv 0 \pmod{43}$ $4k^2 + 4k + 4 \equiv 0 \pmod{43}$ $(2k + 1)^2 + 3 \equiv 0 \pmod{43}$ $(2k + 1)^2 \equiv -3 \equiv 169 = 13^2 \pmod{43}$

Taking square roots modulo $43$:

1. $2k + 1 \equiv 13 \pmod{43} \implies 2k \equiv 12 \pmod{43} \implies k \equiv 6 \pmod{43}$
2. $2k + 1 \equiv -13 \equiv 30 \pmod{43} \implies 2k \equiv 29 \equiv 72 \pmod{43} \implies k \equiv 36 \pmod{43}$

The smallest integer value of $k \in \mathbb{N}$ for $k > 1$ is $p = 6$.

Now, calculate $n$: $n = \frac{p^3 - 1}{43} = \frac{6^3 - 1}{43} = \frac{215}{43} = 5 \in \mathbb{N}$

Step 5: Find $p + n$

$p + n = 6 + 5 = 11$

Correct Option: B