Key Concepts and Formulas

This problem primarily relies on the properties of binomial coefficients, specifically the identity that allows us to simplify expressions involving $r \cdot {^n C_r}$. The two crucial identities are:

1. Identity for $r \cdot {^n C_r}$: $r \cdot {^n C_r} = n \cdot {^{n-1} C_{r-1}}$ This identity is derived from the definition of binomial coefficients: $r \cdot \frac{n!}{r!(n-r)!} = r \cdot \frac{n!}{r \cdot (r-1)!(n-r)!} = \frac{n!}{(r-1)!(n-r)!} = n \cdot \frac{(n-1)!}{(r-1)!(n-r)!} = n \cdot {^{n-1} C_{r-1}}$. It's particularly useful when dealing with sums like $\sum r \cdot {^n C_r}$ or $\sum r^2 \cdot {^n C_r}$.

2. Sum of Binomial Coefficients: $\sum_{r=0}^n {^n C_r} = {^n C_0} + {^n C_1} + \ldots + {^n C_n} = 2^n$ This represents the total number of subsets of a set with $n$ elements.

Step-by-Step Solution

We are asked to evaluate the sum $S = 1^2 \cdot\left({ }^{15} C_1\right)+2^2 \cdot\left({ }^{15} C_2\right)+3^2 \cdot\left({ }^{15} C_3\right)+\ldots+15^2 \cdot\left({ }^{15} C_{15}\right)$. This can be written in summation notation as: $S = \sum_{r=1}^{15} r^2 \cdot{ }^{15} C_r$ Note that the sum starts from $r=1$ because for $r=0$, the term $r^2 \cdot {^{15} C_0}$ would be $0^2 \cdot 1 = 0$, so including it or excluding it does not change the sum.

Step 1: Apply the Identity $r \cdot {^n C_r} = n \cdot {^{n-1} C_{r-1}}$

We can rewrite $r^2 \cdot {^{15} C_r}$ as $r \cdot (r \cdot {^{15} C_r})$. Using the identity $r \cdot {^n C_r} = n \cdot {^{n-1} C_{r-1}}$ with $n=15$, we get: $r \cdot {^{15} C_r} = 15 \cdot {^{14} C_{r-1}}$.

Substitute this back into the sum: $S = \sum_{r=1}^{15} r \cdot (15 \cdot {^{14} C_{r-1}})$ We can take the constant $15$ outside the summation: $S = 15 \sum_{r=1}^{15} r \cdot {^{14} C_{r-1}}$

Step 2: Express $r$ as $(r-1)+1$ and Split the Sum

To further simplify the term $r \cdot {^{14} C_{r-1}}$, we use a common technique: express $r$ in terms of $(r-1)$, matching the lower index of the binomial coefficient. $r = (r-1) + 1$ Substitute this into the sum: $S = 15 \sum_{r=1}^{15} ((r-1) + 1) \cdot {^{14} C_{r-1}}$ Now, split the sum into two parts: $S = 15 \left[ \sum_{r=1}^{15} (r-1){ }^{14} C_{r-1} + \sum_{r=1}^{15} { }^{14} C_{r-1} \right]$

Step 3: Evaluate the First Sum: $\sum_{r=1}^{15} (r-1){ }^{14} C_{r-1}$

Let's evaluate the first part of the expression inside the brackets: $\sum_{r=1}^{15} (r-1){ }^{14} C_{r-1}$. Let $j = r-1$. When $r=1$, $j=0$. When $r=15$, $j=14$. The sum transforms to: $\sum_{j=0}^{14} j \cdot {^{14} C_j}$ The term for $j=0$ is $0 \cdot {^{14} C_0} = 0$, so we can start the sum from $j=1$: $\sum_{j=1}^{14} j \cdot {^{14} C_j}$ Now, apply the identity $j \cdot {^N C_j} = N \cdot {^{N-1} C_{j-1}}$ with $N=14$: $= \sum_{j=1}^{14} 14 \cdot {^{13} C_{j-1}}$ Take the constant $14$ outside the sum: $= 14 \sum_{j=1}^{14} {^{13} C_{j-1}}$ Again, let $k = j-1$. When $j=1$, $k=0$. When $j=14$, $k=13$. The sum becomes: $= 14 \sum_{k=0}^{13} {^{13} C_k}$ Using the identity $\sum_{k=0}^N {^N C_k} = 2^N$ with $N=13$: $= 14 \cdot 2^{13}$

Step 4: Evaluate the Second Sum: $\sum_{r=1}^{15} { }^{14} C_{r-1}$

Now, let's evaluate the second part inside the brackets: $\sum_{r=1}^{15} { }^{14} C_{r-1}$. Let $j = r-1$. When $r=1$, $j=0$. When $r=15$, $j=14$. The sum transforms to: $\sum_{j=0}^{14} {^{14} C_j}$ Using the identity $\sum_{j=0}^N {^N C_j} = 2^N$ with $N=14$: $= 2^{14}$

Step 5: Combine the Results

Substitute the results from Step 3 and Step 4 back into the expression for $S$: $S = 15 \left[ (14 \cdot 2^{13}) + (2^{14}) \right]$ To simplify, factor out the common term $2^{13}$: $S = 15 \left[ 14 \cdot 2^{13} + 2 \cdot 2^{13} \right]$ $S = 15 \cdot 2^{13} (14 + 2)$ $S = 15 \cdot 2^{13} \cdot 16$ Since $16 = 2^4$: $S = 15 \cdot 2^{13} \cdot 2^4$ $S = 15 \cdot 2^{17}$

Step 6: Prime Factorization and Final Answer

We need to express $S$ in the form $2^m \cdot 3^n \cdot 5^k$. Prime factorize $15$: $15 = 3 \cdot 5$. So, $S = (3 \cdot 5) \cdot 2^{17}$ $S = 2^{17} \cdot 3^1 \cdot 5^1$ Comparing this with $2^m \cdot 3^n \cdot 5^k$, we identify the exponents: $m = 17$ $n = 1$ $k = 1$

The question asks for the value of $m+n+k$: $m+n+k = 17+1+1 = 19$

Tips and Common Mistakes

• Understanding Binomial Identities: Memorize or be able to quickly derive key binomial identities like $r \cdot {^n C_r} = n \cdot {^{n-1} C_{r-1}}$. These are fundamental for simplifying such sums.
• Adjusting Summation Limits: When performing substitutions like $j=r-1$ or $k=j-1$, always remember to adjust the limits of the summation accordingly. A common error is forgetting to change the lower or upper bounds, which can lead to incorrect results.
• Careful with Algebra: Pay close attention to factoring out common terms and simplifying powers of two. Small arithmetic errors can lead to the wrong final answer.
• Recognizing $\sum {^n C_r} = 2^n$: This identity is frequently used in binomial theorem problems. Ensure you can apply it correctly.

Summary and Key Takeaway

This problem demonstrates a classic approach to evaluating sums involving $r^2 \cdot {^n C_r}$ by iteratively applying the identity $r \cdot {^n C_r} = n \cdot {^{n-1} C_{r-1}}$ and judiciously splitting terms. The key is to reduce the power of $r$ by repeated application of the identity until the sum can be evaluated using the basic sum of binomial coefficients, $\sum {^n C_r} = 2^n$. The final value of $m+n+k$ is $19$.