Problem 58 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The value of \(\sum_{\mathrm{r}=1}^{15} \mathrm{r}^{2}\left(\frac{{ }^{15} \mathrm{C}_{\mathrm{r}}}{{ }^{15} \mathrm{C}_{\mathrm{r}-1}}\right)\) is equal to : (a) 1240 (b) 560 (c) 1085 (d) 680

Step-by-Step Solution

Answer

The value of the expression is 1800.

Step 1: Understand the Formula

The given expression is:\[\sum_{r=1}^{15} r^2 \left(\frac{{ }^{15}C_{r}}{{ }^{15}C_{r-1}}\right)\]The formula for a combination \( ^nC_r \) is \( \frac{n!}{r!(n-r)!} \). Our task is to simplify this expression using these formula properties.

Step 2: Simplify the Combination Ratio

First, simplify the combination ratio by using the formula for combination:\[\frac{{ }^{15}C_{r}}{{ }^{15}C_{r-1}} = \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(15-(r-1))!}}\]This simplifies to:\[\frac{15!(15-r+1)}{r!(r-1)!(15-r)!} = \frac{15}{r}\]

Step 3: Substitute Simplified Ratio

Now substitute the simplified ratio back into the series:\[\sum_{r=1}^{15} r^2 \left(\frac{15}{r}\right) = 15 \sum_{r=1}^{15} r\]

Step 4: Calculate the Sum of r

The sum of the first 15 natural numbers \( \sum_{r=1}^{15} r \) is calculated using the formula for the sum of an arithmetic series:\[\sum_{r=1}^{15} r = \frac{15(15+1)}{2} = 120\]

Step 5: Multiply to Find the Answer

Multiply the sum by 15:\[15 \times 120 = 1800\]

Key Concepts

Combinatorics

Combinatorics is a fascinating area of mathematics that deals with counting, arrangement, and combination of elements in sets. It's like solving puzzles by figuring out how many ways you can organize things. In this exercise, we specifically look at combinations, which is selecting items from a group where the order doesn’t matter. For example, choosing 2 fruits from a basket of 3 bananas and 2 apples calculating combinations is key.The formula used is the combination formula, denoted as \(^nC_r\) and expressed as \(\frac{n!}{r!(n-r)!}\). Here, \(n!\) means factorial, which is the product of all positive integers less than or equal to \(n\). Understanding this formula is critical to solve many problems in combinatorics that involve counting different possible scenarios without arranging them.

Arithmetic Series

An arithmetic series is a sequence of numbers in which each term after the first is derived by adding a constant to the previous one. This constant is called the "common difference."In this exercise, the common difference is \(1\), as we are summing numbers from 1 to 15. The task was to find the sum of the first 15 integers.The formula for the sum \(S\) of an arithmetic series for \(n\) terms is:

• \(S = \frac{n}{2} (a_1 + a_n)\)

where \(a_1\) is the first term, and \(a_n\) is the last term. For our arithmetic series:

• \(\sum_{r=1}^{15} r = \frac{15 \times (1 + 15)}{2} = 120\)

This formula simplifies the process of adding a sequence of increments, making calculations like this swift and systematic.

Binomial Coefficient

The binomial coefficient, often shown as \( ^nC_r\), is a central part of combinatorics. It appears in many areas of mathematics, including algebra and probability. It's used to find the number of ways to choose \(r\) elements from a set of \(n\) elements.In the exercise, the binomial coefficient was simplified through division \(\frac{15}{r}\). This simplification is crucial because it helps us manage series and sequences more easily. By breaking down the expression, we were able to focus on multiplying sequences instead of handling complex divisions. The binomial coefficient can also be visualized in Pascal’s Triangle, which illustrates how each number is built on the sum of two directly above it. This helps us understand relationships within combinations and solve more complex problems.