Problem 311 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Sum the infinite series \(\frac{12}{2 !}+\frac{28}{3 !}+\frac{50}{4 !}+\frac{78}{5 !}+\cdots\)

Step-by-Step Solution

Answer

So, the sum of the given infinite series is \(6e^1\), where \(e\) is the base of natural logarithm.

Step 1: Identify patterns

First, we can clearly observe a pattern here. In the numerator, it seems like we are adding 16 for each next term: 12, 28, 50, 78... More specifically, it can be written as 2n(n+2) where n>=1. In the denominator, we have factorial numbers starting from 2: 2!, 3!, 4!, 5!... This can be expressed as (n+1)!. So each term of this series can be represented as \(\frac{2n(n+2)}{(n+1)!}\) where n is from 1 to infinity.

Step 2: Rewrite the series as a combination of two series

Notice that \(\frac{2n(n+2)}{(n+1)!}\) can be rewritten as \(\frac{2n^2+4n}{(n+1)!} = \frac{2n^2}{(n+1)!} + \frac{4n}{(n+1)!}\), we can see that each of the two terms can generate a series when summed from n=1 to infinity. The given series can thus be rewritten as sum of two series: \(\sum_{n=1}^{\infty} \frac{2n^2}{(n+1)!}+\sum_{n=1}^{\infty} \frac{4n}{(n+1)!}\)

Step 3: Further simplify each series

The first series is equal to \(2\sum_{n=1}^{\infty} \frac{n}{n!} = 2\sum_{n=1}^{\infty} \frac{1}{(n-1)!}\), which is 2 times the sum of \(e^1\) from n=1 to infinity. Similarly, the second series equals \(4\sum_{n=1}^{\infty} \frac{1}{(n)!}\), which sums up to \(4e^1\) from n=1 to infinity.

Step 4: Add up the sums

We have separated our sequence into two series both of which sum up to \(e^1\). Each of those sums are multiplied by 2 and 4 respectively, thus, we add them together to get our final answer: \(2(e^1) + 4(e^1) = 6e^1\)

Key Concepts

Factorial

Factorial is a mathematical function often represented by the symbol "!". It is used to multiply a series of descending natural numbers. For example, the factorial of 3, denoted as 3!, is calculated as:\[ 3! = 3 \times 2 \times 1 = 6 \]In this case, the term (n+1)! describes a sequence of products:

• For 2!, it is 2.
• For 3!, it becomes 3 × 2.
• For 4!, it is 4 × 3 × 2.

This pattern of multiplication continues indefinitely, and it is fundamental in formulating expressions in series. Factorials grow rapidly with larger numbers and are crucial in calculations involving permutations, combinations, and in this exercise, series expansions.

Series Expansion

Series expansion is a way we express a mathematical function as a sum of terms calculated from the values of its derivatives at a single point. Often, the series expansion involves terms like fractions and factorials, as seen in the exercise.The original series given as:\[ \frac{12}{2!}+\frac{28}{3!}+\frac{50}{4!}+\cdots \]illustrates how a function can expand into an infinite series. By identifying the formula for each term, we write:\[ \frac{2n(n+2)}{(n+1)!} \]where "n" represents any integer greater than or equal to 1.

• This gives insight into how coefficients and factorials create a repeating pattern in an infinite sequence.
• Such expansions are powerful in approximating functions and calculating certain sums, especially when we transform them into recognizable mathematical constants like the exponential function.

Exponential Function

The exponential function, denoted as \( e^x \), is a crucial component in mathematical analysis and is instrumental to series like the one in this exercise. The number \( e \) is a mathematical constant approximately equal to 2.71828, and it serves as the base for natural logarithms.In the context of series:

• The exponential function plays a key role when dealing with series whose terms can be rewritten involving factorials, as it helps to sum infinite series effectively.
• For example, the series:\[ \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \]expresses part of the process of calculating powers of \( e \).

Understanding the exponential function helps in summing infinite series efficiently, as shown in the step-by-step solution where each part of the series is linked back to \( e^1 \). This facilitates the simplification and calculation of complex series in a very elegant way.