If 1^2 * (15C1) + 2^2 * (15C2) + 3^2 * (15C3) + ... + 15^2 * (15C15) = 2^m * 3^n * 5^k, where m, n,

Q: Why can't we just use the binomial expansion $(1+x)^n$ directly?

While you could differentiate $(1+x)^n$ twice to get $r^2$ terms, it becomes algebraically messy with the $x=1$ substitution. The algebraic decomposition method $r^2 = r(r-1) + r$ is much faster and less prone to calculation errors in a JEE exam setting.

Q: Why does the first sum start at $r=2$?

Because for $r=0$ and $r=1$, the term $r(r-1)$ is zero. Since these terms contribute nothing to the sum, we can safely start the summation index from $r=2$ to make the binomial identity application cleaner.

Q: How do we know $n(n-1) \binom{n-2}{r-2}$ is correct?

You can verify this by expanding the combinations into factorials: $r(r-1) \frac{n!}{r!(n-r)!} = \frac{n!}{(r-2)!(n-r)!} = n(n-1) \frac{(n-2)!}{(r-2)!(n-r)!} = n(n-1) \binom{n-2}{r-2}$. It is a standard reduction technique in combinatorics.

Question

If $1^2 \cdot ({}^{15}C_1) + 2^2 \cdot ({}^{15}C_2) + 3^2 \cdot ({}^{15}C_3) + \dots + 15^2 \cdot ({}^{15}C_{15}) = 2^m \cdot 3^n \cdot 5^k$, where $m, n, k \in \mathbb{N}$, then $m + n + k$ is equal to :

Options:

(A) 19

(B) 21

(C) 18

(D) 20

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Series

Welcome, student. Today, we are going to peel back the layers of a classic JEE Advanced problem.

At first glance, the series $S = 1^2 \binom{15}{1} + 2^2 \binom{15}{2} + \dots + 15^2 \binom{15}{15}$ might look like a daunting wall of numbers. But in the world of competitive mathematics, such series are not walls; they are puzzles waiting for the right key.

Our goal is to evaluate this sum and express it as $2^m \cdot 3^n \cdot 5^k$. Let us embark on this journey together.

### The Obstacle: The $r^2$ Term

The core difficulty here is the $r^2$ coefficient. If it were just $\binom{n}{r}$, we would know the answer is $2^n$.

If it were $r \binom{n}{r}$, we would use the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$. But $r^2$ is a different beast.

To tame it, we use a clever algebraic trick. We rewrite $r^2$ as $r(r-1) + r$. This decomposition allows us to split the summation into two distinct parts, each of which can be solved using the binomial identity mentioned above.

### The Strategy: Splitting the Sum

By substituting our decomposition into the summation, we get:

$$S = \sum_{r=1}^{n} [r(r-1) + r] \binom{n}{r}$$

We can now split this into two separate sums:

$$S = \sum_{r=1}^{n} r(r-1) \binom{n}{r} + \sum_{r=1}^{n} r \binom{n}{r}$$

Notice that for the first part, when $r=0$ or $r=1$, the term $r(r-1)$ becomes zero. This is a gift! It means we can start our index from $r=2$ without changing the value of the sum, making the math much cleaner.

### The Execution: Applying Identities

Now, we apply our binomial identity. For the first part, $\sum_{r=2}^{n} r(r-1) \binom{n}{r}$, we use the identity twice to pull out $n(n-1)$, leaving us with:

$$\sum_{r=2}^{n} r(r-1) \frac{n!}{r!(n-r)!} = n(n-1) \sum_{r=2}^{n} \binom{n-2}{r-2} = n(n-1) 2^{n-2}$$

For the second part, $\sum_{r=1}^{n} r \binom{n}{r}$, we use the identity once to get:

$$n \sum_{r=1}^{n} \binom{n-1}{r-1} = n 2^{n-1}$$

Combining these, we get the general form:

$$S = n(n-1) 2^{n-2} + n 2^{n-1}$$

### The Final Simplification

Let us factor out the common terms. Factoring out $n 2^{n-2}$, we are left with:

$$S = n 2^{n-2} [ (n-1) + 2 ] = n(n+1) 2^{n-2}$$

Now, we substitute $n=15$:

$$S = 15(16) 2^{13}$$

Since $16 = 2^4$, this becomes $15 \cdot 2^4 \cdot 2^{13} = 15 \cdot 2^{17}$. Breaking $15$ into its prime factors, $3^1 \cdot 5^1$, we arrive at:

$$S = 2^{17} \cdot 3^1 \cdot 5^1$$

Comparing this to $2^m \cdot 3^n \cdot 5^k$, we find $m=17, n=1, k=1$. The final result is **$m+n+k = 17+1+1 = 19$**. You have just conquered a complex series with elegance and logic.

Animated Solution for Mathematics - Binomial Theorem: If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :

Select Answer:

Visualized Solution

Identifying the Series Pattern

The Core Binomial Identity

Splitting the Summation

Applying Identities Twice

Evaluating the Sums

Combining the Results

Substituting $n = 15$

Prime Factorization

Final Calculation of $m + n + k$

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Binomial Series

The Obstacle

The Strategy

The Execution

The Final Simplification

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