Problem 68 - A Complete Resource Book in Mathematics for JEE Main

Question

Sum to \(n\) terms of the series \(\frac{1}{5 !}+\frac{1 !}{6 !}+\frac{2 !}{7 !}+\frac{3 !}{8 !}+\ldots\) is (A) \(\frac{2}{5 !}-\frac{1}{(n+1) !}\) (B) \(\frac{1}{4}\left(\frac{1}{4 !}-\frac{n !}{(n+4) !}\right)\) (C) \(\frac{1}{4}\left(\frac{1}{3 !}-\frac{3 !}{(n+2) !}\right)\) (D) None of these

Step-by-Step Solution

Answer

The answer is (B) \(\frac{1}{4}\left(\frac{1}{4!}-\frac{n!}{(n+4)!}\right)\).

Step 1: Identify the General Term of the Series

To find the general term of the series, we notice the numerators follow the pattern of factorials starting from 0, while the denominators follow one factorial larger than the numerators. Thus, the nth-term of the series can be expressed as: \( \frac{(n-1)!}{(n+4)!} \).

Step 2: Express the Series Sum

We are tasked to find the sum of \( n \) terms of the series. Therefore, we sum over the expression for the general term as \( \sum_{r=1}^{n} \frac{(r-1)!}{(r+4)!} \).

Step 3: Simplify the General Term

Write the general term \( \frac{(n-1)!}{(n+4)!} \) in a simplified way to identify possible telescopic summation. Use factorial identities: \( \frac{(n-1)!}{(n+4)!} = \frac{1}{(n)(n+1)(n+2)(n+3)(n+4)} \).

Step 4: Apply Telescoping Method

Observe that part of the general term pattern can be expressed as a telescoping series. Break the term down as a series of partial fractions that reveal the telescoping nature. Decomposing: \[ \frac{1}{r(r+1)(r+2)(r+3)(r+4)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2} + \frac{D}{r+3} + \frac{E}{r+4} \] and solve for A, B, C, D, and E.

Step 5: Evaluate the Sum

Summing these terms from \( r = 1 \) to \( r = n \), after partial fraction decomposition, leads to a telescoping series where most terms cancel out, leaving:\[ \sum_{r=1}^{n} \left( \frac{1}{4}\frac{1}{4!} - \frac{(n-1)!}{(n+4)!} \right) \].

Step 6: Substitute and Simplify

Simplify the result from the telescoping series to derive the formula. The surviving terms yield: \[ \text{Sum} = \frac{1}{4}\left(\frac{1}{4!} - \frac{n!}{(n+4)!}\right) \]. This matches option (B).

Key Concepts

Factorials

Factorials are a fundamental concept in mathematics, often represented by the symbol "!". They are used to express the product of an integer and all the integers below it. For instance, the factorial of 5 is written as 5! and is calculated as 5 x 4 x 3 x 2 x 1 = 120. Factorials appear frequently in arrangements or combinations, where ordering of items matters.

• The factorial function increases very rapidly with larger numbers, making it useful for diverse applications where permutations and combinations need to be counted.
• Factorials are foundational in deriving the terms in series such as the one in the original problem, where both numerator and denominator consist of factorial expressions.

Understanding the growth and structure of factorials is crucial in manipulating and simplifying series expressions, such as converting them into forms suitable for telescoping or other summation techniques.

Telescoping Series

A telescoping series is a sequence of terms where consecutive terms cancel each other out significantly during summation. This makes the series much easier to evaluate because most terms disappear through successive partial cancellations.

• When you break a formula into smaller parts using partial fractions, you reveal a pattern where terms from one fraction will cancel terms from another, leaving only the first and last few terms.
• In the original problem, the series sum used a telescoping technique to simplify the otherwise complex addition of terms using a strategic breakdown, dramatically easing the calculation process.

Recognizing a telescoping nature means analyzing the differences often hidden within the series, allowing us to apply the method effectively and achieve a compact solution.

Partial Fractions

Partial fraction decomposition is a method used to express a rational function as the sum of simpler fractions. This technique is vital for simplifying complex algebraic expressions, particularly those involving polynomials.

• To apply partial fractions, first, ascertain that the problem can be split into terms of smaller degree polynomials in the denominator. These forms are more manageable and often reveal cancellations.
• In our series, partial fractions helped decompose the problem into separate components, creating an opportunity to see how the series telescopes. Each fraction component matched a part of the original denominator, allowing for specific numerators (A, B, C, D, E) to be identified and solved.

This approach highlights the power of partial fractions in mathematical problem-solving, making complicated summations into approachable tasks.