Question

If coefficient of $x^3$ in $(1+x)^3 + (1+x)^4 + \dots + (1+x)^{99} + (1+kx)^{100}$ is $\binom{100}{3} \left( \frac{101}{4} - 43n \right)$, then the value of $(k^3 + 43n)$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The sum $S = (1+x)^3 + (1+x)^4 + \dots + (1+x)^{99}$ is a GP with common ratio $(1+x)$.
Coefficient of $x^r$ in $(1+x)^n$ is $\binom{n}{r}$.
Step 2: Detailed Explanation:
1. Sum of GP:
$S = \frac{(1+x)^3 [ (1+x)^{97} - 1 ]}{(1+x) - 1} = \frac{(1+x)^{100} - (1+x)^3}{x}$.
2. Coefficient of $x^3$ in $S$:
This is the coefficient of $x^4$ in $(1+x)^{100} - (1+x)^3$, which is $\binom{100}{4} - 0 = \binom{100}{4}$.
3. Total coefficient of $x^3$:
$\binom{100}{4} + \binom{100}{3} k^3$.
Equating to $\binom{100}{3} [ \frac{101}{4} - 43n ]$:
$\frac{100 \cdot 99 \cdot 98 \cdot 97}{4 \cdot 3 \cdot 2 \cdot 1} + \binom{100}{3} k^3 = \binom{100}{3} [ \frac{101}{4} - 43n ]$.
Using $\binom{100}{4} = \binom{100}{3} \cdot \frac{97}{4}$:
$\frac{97}{4} + k^3 = \frac{101}{4} - 43n$.
$k^3 + 43n = \frac{101}{4} - \frac{97}{4} = 1$.
Step 4: Final Answer:
The value is 1.