Question

If
$ 2^m 3^n 5^k, \text{ where } m, n, k \in \mathbb{N}, \text{ then } m + n + k \text{ is equal to:} $

• A. 19
• B. 21
• C. 18
• D. 20

Hint:

In series problems involving binomial coefficients, use the binomial expansion and properties of the binomial sum to simplify the expression.

Answer 1

The Correct Option is A

The given series is: \[ \sum_{r=1}^{15} 2^r \binom{15}{r} \] This can be rewritten as: \[ \sum_{r=1}^{15} r \binom{15}{r-1} \] Now, compute this using the binomial expansion formula. The expression simplifies to: \[ 15 \times 14 \times 2^{13} \quad \text{(binomial expansion terms)} \] Substituting the values into the sum: \[ 15 \times 14 \times 2^{13} = 15 \times 14 \times 2^{14} \quad \Rightarrow \quad m = 17, n = 1, k = 1 \]
Thus, \( m + n + k = 17 + 1 + 1 = 19 \).
Thus, the correct answer is \( 19 \).

Answer 2

Step 1: The given summation is: 

\[ \sum_{r=1}^{15} r^2 \binom{15}{r} \Rightarrow 15 \sum_{r=1}^{15} r^{14} \binom{r-1}{1} \]

Step 2: Simplifying the summation:

\[ 15 \sum_{r=1}^{15} (r - 1 + 1)^{14} \binom{r-1}{1} \] which further simplifies to: \[ 15 \cdot 14 \sum_{r=1}^{15} \binom{r-2}{13} + 15 \cdot 14 \sum_{r=1}^{15} \binom{r-1}{14}. \]

Step 3: Calculating the terms:

\[ 15 \cdot 14 \cdot 2^{13} + 15 \cdot 14 \cdot 2^{14}. \]

Step 4: Final Simplification:

This can be simplified as: \[ 3^1 \cdot 2^{13} (70 + 10) = 3^1 \cdot 2^{13} \cdot 80. \]

Step 5: Final Calculation:

Further simplifying: \[ 3^1 \cdot 5^1 \cdot 2^{17}. \]

Step 6: Final Result:

\[ m = 17n = 1k = 1. \]

\(m+n+k = 17+1+1 = 19\)