Problem 31 - 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

Option C: \( \frac{1}{4}\left(\frac{1}{3!}-\frac{3!}{(n+2)!}\right) \).

Step 1: Understand the general term of the series

The given series is \( \frac{1}{5!} + \frac{1!}{6!} + \frac{2!}{7!} + \frac{3!}{8!} + \ldots \). We identify the pattern for the general term, \( a_n \), as \( \frac{(n-1)!}{(n+4)!} \).

Step 2: Simplify the general term

We have \( a_n = \frac{(n-1)!}{(n+4)!} = \frac{1}{(n+4)(n+3)(n+2)(n+1) n (n-1) (n-2) \ldots 1 } \) with \( (n-1)! \) terms canceling some terms in the denominator.

Step 3: Check for telescopic nature

Notice that \( a_n \) can be expressed as \( \frac{1}{4} \left( \frac{1}{n+3!} - \frac{n!}{(n+4)!} \right) \) by expanding further and recognizing any cancellations possible, embodying a telescopic nature.

Step 4: Calculate the sum to n terms

The series' sum, denoted by \( S_n \), therefore becomes a telescopic series, \( S_n = \sum_{n=1}^{n} \frac{1}{4} \left( \frac{1}{3!} - \frac{n!}{(n+4)!} \right) \). The sums telescope, leading us to find the end terms factor out remaining as \( \frac{1}{4} \left( \frac{1}{3!} - \frac{n!}{(n+4)!} \right) \).

Step 5: Identify matching answer

Compare with the options given. The sum is \( \frac{1}{4} \left( \frac{1}{3!} - \frac{3!}{(n+2)!} \right) \), which corresponds to option C.

Key Concepts

Sequence and Series

Understanding the concepts of sequences and series is crucial in mathematics. A sequence is a list of numbers arranged in a specific order, and each number in the list is called a term. A series, on the other hand, represents the sum of the elements of a sequence. For instance, in a series like \[\frac{1}{5!} + \frac{1!}{6!} + \frac{2!}{7!} + \cdots\]each fraction is a term, and adding them up gives us a series.
Telescoping series is a unique type of series where consecutive terms cancel out each other. This property simplifies finding the sum significantly since most of the intermediate terms vanish when the series is summed. This leaves us only with the first and last few terms, making the calculation easier. In the exercise, the series is telescopic, which facilitates the simplification of the series' sum.

Factorial Notation

Factorial notation is a mathematical concept expressed as \( n! \) which represents the product of all positive integers up to \( n \). For example, \( 5! \) equals \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \). This notation is often used in permutations, combinations, and series computations.
In the series given in the exercise, factorial notation helps provide a structured way to express terms. Each term is a fraction where both the numerator and the denominator are factorials, like \( \frac{(n-1)!}{(n+4)!} \). The factorial in the denominator grows much faster than in the numerator, hence each term of the series becomes smaller as \( n \) increases.

Partial Fraction Decomposition

Partial fraction decomposition is a technique in which a complex rational expression is expressed as a sum of simpler fractions. This is especially useful in integration and summation of series where direct calculation is challenging.
For the exercise, partial fraction decomposition is used to break down the term \( \frac{(n-1)!}{(n+4)!} \) into a simpler form. This decomposed format allows us to identify cancellation terms easily, making the series telescopic.

• Partial fraction decomposition simplifies complex numerator/denominator relationships.
• It helps find a pattern of cancellation in telescoping series.
• Provides a clearer structure when summing series regularly or to the limit of infinity.

These decompositions were key to identifying the telescopic nature of the series in the exercise. This led to a straightforward summation result and matched perfectly with one of the provided solution options.