Problem 53 - A Complete Resource Book in Mathematics for JEE Main

Question

\({ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left[{ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}} C_{2}\right]=\) (A) \(\frac{n(n+1)(2 n+1)}{6}\) (B) \(\frac{n(n+1)}{2}\) (C) \(\frac{n(n-1)(2 n-1)}{6}\) (D) none of these

Step-by-Step Solution

Answer

(A) \(\frac{n(n+1)(2n+1)}{6}\).

Step 1: Understand the Problem

We need to solve the given expression and simplify it. The expression involves binomial coefficients, specifically the combinations of selecting 2 items from n+1, 2, 3, ..., n items.

Step 2: Simplify Binomial Coefficients

Recall that the binomial coefficient \({ }^{r}C_{2}\) is given by \(\frac{r(r-1)}{2}\). We apply this formula to each term in the sum: \[{ }^{2}C_{2} = 1, \ { }^{3}C_{2} = 3, \ { }^{4}C_{2} = 6, \ldots, {}^{\text{n}}C_{2} = \frac{n(n-1)}{2}\]

Step 3: Evaluate the Summation in Brackets

We need to compute the sum \(2[{ }^{2}C_{2} + { }^{3}C_{2} + \ldots + { }^{n}C_{2}]\). Using the formula for each \({ }^{r}C_{2}\):\[ \sum_{r=2}^{n}\binom{r}{2} = \sum_{r=2}^{n}\frac{r(r-1)}{2}.\] This becomes \[\frac{1}{2}[1 + 3 + 6 + \ldots + \frac{n(n-1)}{2}]\].

Step 4: Replace with Known Formula

The sum of the first k natural numbers is \(\frac{k(k+1)}{2}\) and the sum of the first k squares is \(\frac{k(k+1)(2k+1)}{6}\). We aim for such a form to simplify this sum. \[\sum_{r=2}^{n} r^2 - r = \{\sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r\} + (1 - 1) = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}\].

Step 5: Complete the Simplification

Finish simplification by evaluating the expression: \[ 2\left(\sum_{r=2}^{n} \frac{r(r-1)}{2}\right) = 2\left(\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}\right) \]. We now substitute \( { }^{n+1}C_{2} = \frac{(n+1)n}{2} \) and combine the terms.

Step 6: Combine Terms

Subsequently solve: \[\frac{(n+1)n}{2} + \frac{n(n+1)(2n+1)}{3} - n(n+1)\]. Simplifying using a common denominator gives us \[\frac{n(n+1)(2n+1) + 3(n)(n+1) - 6n(n+1)/6 = \frac{n(n+1)(2n+1)}{6}\] which corresponds to choice (A).

Key Concepts

Combinatorial Mathematics

Combinatorial mathematics is a fascinating branch of mathematics focused on counting, arrangement, and combination of elements within a set. It is the backbone of many mathematical concepts that require a deeper understanding of how individual items can be grouped together. One of the prominent tools in combinatorial mathematics is the binomial coefficient, also known as the combination.The binomial coefficient \({ }^{n}\mathbb{C}_k\) represents the number of ways to choose \(k\) elements from a set of \(n\) elements without regard to the order of selection. The formula for this is given by:

• \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

This formula allows us to calculate how many different groups of \(k\) we can form from \(n\) elements. It's used not only in algebra but also in numerous fields like statistics, probability theory, and computer science. Understanding and applying these coefficients help in simplifying expressions, solving combinatorial problems, and finding specific summations.

Summation Formulas

In mathematics, summation formulas are essential for evaluating the sum of sequences or series. When dealing with expressions involving binomial coefficients, understanding summation is critical.The specific problem involves the series \( \sum_{r=2}^{n} \binom{r}{2} \), where each term represents the different combinations of choosing 2 items from \(r\) items. This can be further broken down using its formula \( \binom{r}{2} = \frac{r(r-1)}{2} \), simplifying the summation expression to:

• \[ \sum_{r=2}^{n} \frac{r(r-1)}{2} \]

Working with summation formulas is crucial because they allow us to substitute well-known series, such as the sum of natural numbers and the sum of squares, into our calculations. By doing so, difficult problems become manageable, reducing complex expressions into simpler ones we can work with more easily. Through this problem, we also explore the use of known formulas like:

• Sum of first \(k\) natural numbers: \( \frac{k(k+1)}{2} \)
• Sum of first \(k\) squares: \( \frac{k(k+1)(2k+1)}{6} \)

Algebraic Simplification

Algebraic simplification is the process of making a mathematical expression more manageable and easier to solve. It often involves reducing the number of terms, simplifying fractions, and canceling common factors.In our exercise, we used algebraic simplification to tackle the complex expression involving binomial coefficients. After applying the respective known summation formulas, we arrive at the expression:

• \[ 2 \left( \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \right) \]

The solution then further simplifies to:

• \[ \frac{(n+1)n}{2} + \frac{n(n+1)(2n+1)}{3} - n(n+1) \]

This is completed by combining terms under a common denominator, leading to the final form which represents choice (A):

• \[ \frac{n(n+1)(2n+1)}{6} \]

Algebraic simplification helps us bridge the gap from a cumbersome expression to an elegant solution, demonstrating the power of algebra to resolve complex problems smoothly and accurately.