Question

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

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given a series of binomial expansions and asked to find the coefficient of \(x^3\).
The coefficient of \(x^3\) in \((1+ax)^m\) is \(\binom{m}{3}a^3\).
Step 2: Key Formula or Approach:
The given expression is \(S = \sum_{r=3}^{99} (1+x)^r + (1+kx)^{100}\).
The coefficient of \(x^3\) in \(\sum_{r=3}^{99} (1+x)^r\) is \(\sum_{r=3}^{99} \binom{r}{3}\).
Using the hockey-stick identity, \(\sum_{r=m}^{n} \binom{r}{m} = \binom{n+1}{m+1}\), we have:
\[ \sum_{r=3}^{99} \binom{r}{3} = \binom{100}{4} \] The coefficient of \(x^3\) in \((1+kx)^{100}\) is \(\binom{100}{3}k^3\).
So, the total coefficient of \(x^3\) is \(\binom{100}{4} + \binom{100}{3}k^3\).
Step 3: Detailed Explanation:
According to the problem, this coefficient is equal to \(\left( 43n + \frac{101}{4} \right) \binom{100}{3}\):
\[ \binom{100}{4} + \binom{100}{3}k^3 = \left( 43n + \frac{101}{4} \right) \binom{100}{3} \] Divide the entire equation by \(\binom{100}{3}\):
\[ \frac{\binom{100}{4}}{\binom{100}{3}} + k^3 = 43n + \frac{101}{4} \] We compute the ratio:
\[ \frac{\binom{100}{4}}{\binom{100}{3}} = \frac{\frac{100!}{4! \cdot 96!}}{\frac{100!}{3! \cdot 97!}} = \frac{97}{4} \] Substitute this back into the equation:
\[ \frac{97}{4} + k^3 = 43n + \frac{101}{4} \implies k^3 = 43n + \frac{101 - 97}{4} = 43n + 1 \] This implies \(k^3 - 1 = 43n\), meaning \(k^3 \equiv 1 \pmod{43}\).
Since \(n \in \mathbb{N}\), \(n \ge 1\), which means \(k^3 - 1 \ge 43 \implies k>1\).
We test natural numbers \(k>1\) to find the smallest \(k\) satisfying the congruence:
- For \(k = 2\), \(2^3 = 8 \not\equiv 1 \pmod{43}\)
- For \(k = 3\), \(3^3 = 27 \not\equiv 1 \pmod{43}\)
- For \(k = 4\), \(4^3 = 64 = 43(1) + 21 \not\equiv 1 \pmod{43}\)
- For \(k = 5\), \(5^3 = 125 = 43(2) + 39 \not\equiv 1 \pmod{43}\)
- For \(k = 6\), \(6^3 = 216 = 43(5) + 1 \equiv 1 \pmod{43}\)
Thus, the smallest \(k\) is \(6\), which means \(p = 6\).
Step 4: Final Answer:
Substituting \(k = 6\) to find \(n\):
\[ 216 - 1 = 43n \implies 215 = 43n \implies n = 5 \] We are asked to find the value of \(p + n\):
\[ p + n = 6 + 5 = 11 \]