Key Concepts and Formulas

• Binomial Coefficient Ratio: $\frac{{}^n{C_r}}{{}^n{C_{r-1}}} = \frac{n-r+1}{r}$
• Sum of first n natural numbers: $\sum\limits_{r = 1}^{n} {r} = \frac{n(n+1)}{2}$
• Sum of squares of first n natural numbers: $\sum\limits_{r = 1}^{n} {{r^2}} = \frac{n(n+1)(2n+1)}{6}$

Step-by-Step Solution

Step 1: Simplify the Binomial Coefficient Ratio

We are given the expression $\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right)$. The first step is to simplify the ratio of binomial coefficients. Using the formula $\frac{{}^n{C_r}}{{}^n{C_{r-1}}} = \frac{n-r+1}{r}$ with $n=15$, we have: $\frac{{}^{15}{C_r}}{{}^{15}{C_{r-1}}} = \frac{15 - r + 1}{r} = \frac{16 - r}{r}$ Explanation: This simplification is crucial because it transforms the complex binomial coefficient ratio into a simple algebraic expression involving $r$, which is much easier to work with inside a summation.

Step 2: Substitute and Simplify the Term within the Summation

Now, substitute this simplified ratio back into the original summation expression: $S = \sum\limits_{r = 1}^{15} {{r^2}} \left( {\frac{{16 - r}}{r}} \right)$ Next, simplify the term inside the summation: ${r^2} \left( {\frac{{16 - r}}{r}} \right) = r \cdot r \left( {\frac{{16 - r}}{r}} \right)$ Since $r$ starts from $1$, $r \neq 0$, so we can cancel one $r$ from the numerator and denominator: $= r(16 - r)$ $= 16r - r^2$ Explanation: Simplifying the term before performing the summation significantly reduces the complexity. By canceling out $r$ and expanding the product, we get a polynomial in $r$, which is a standard form for applying summation formulas.

Step 3: Deconstruct the Summation

Now our summation becomes: $S = \sum\limits_{r = 1}^{15} {(16r - {r^2})}$ Using the linearity property of summations, we can separate this into two simpler summations: $S = \sum\limits_{r = 1}^{15} {16r} - \sum\limits_{r = 1}^{15} {{r^2}}$ $S = 16\sum\limits_{r = 1}^{15} {r} - \sum\limits_{r = 1}^{15} {{r^2}}$ Explanation: This step is taken to transform the single, combined summation into a combination of standard summations for which we have well-known formulas.

Step 4: Recall and Apply Standard Summation Formulas

We need to calculate the sum of the first $n$ natural numbers and the sum of the squares of the first $n$ natural numbers. For this problem, $n=15$.

The formulas are:

1. Sum of the first $n$ natural numbers: $\sum\limits_{r = 1}^{n} {r} = \frac{n(n+1)}{2}$
2. Sum of the squares of the first $n$ natural numbers: $\sum\limits_{r = 1}^{n} {{r^2}} = \frac{n(n+1)(2n+1)}{6}$

Applying these formulas for $n=15$:

For the first part: $\sum\limits_{r = 1}^{15} {r} = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120$

For the second part: $\sum\limits_{r = 1}^{15} {{r^2}} = \frac{15(15+1)(2(15)+1)}{6} = \frac{15 \times 16 \times (30+1)}{6} = \frac{15 \times 16 \times 31}{6}$ To simplify the calculation, we can divide 15 by 3 and 16 by 2: $= \frac{(3 \times 5) \times (2 \times 8) \times 31}{2 \times 3} = 5 \times 8 \times 31 = 40 \times 31 = \frac{1240}{1}$

Step 5: Final Calculation

Now, substitute the calculated values of the summations back into the expression from Step 3: $S = 16\left(\sum\limits_{r = 1}^{15} {r}\right) - \left(\sum\limits_{r = 1}^{15} {{r^2}}\right)$ $S = 16(120) - 1240$ Perform the multiplication: $16 \times 120 = 1920$ Finally, perform the subtraction: $S = 1920 - 1240 = 680$

Therefore, the value of the given expression is 680.

Common Mistakes & Tips

• Incorrect Application of the Ratio Formula: A very common mistake is to invert the ratio or incorrectly substitute $n$ and $r$. Always remember that $\frac{{}^n{C_r}}{{}^n{C_{r-1}}} = \frac{n-r+1}{r}$.
• Arithmetic Errors: Even with the correct formulas, calculation mistakes can lead to the wrong answer. Double-check all multiplications and subtractions, especially with larger numbers.
• Forgetting Summation Formulas: Memorizing the formulas for $\sum r$ and $\sum r^2$ is essential for JEE. If you don't recall them, derive them quickly or be prepared for a longer solution.

Summary

We simplified the given summation by first using the binomial coefficient ratio formula to reduce the expression inside the summation. Then, we separated the summation into two standard summations: the sum of the first 15 natural numbers and the sum of the squares of the first 15 natural numbers. Finally, we applied the formulas for these standard summations and calculated the result, which is 680.

The final answer is $\boxed{680}$, which corresponds to option (B).