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Question: Find the sum of the series \(\sum\limits_{r=1}^{n}{r\times r!}\)...

Find the sum of the series ∑r=1nr×r!\sum\limits_{r=1}^{n}{r\times r!}

Explanation

Solution

- Hint: First look at the definition of factorial. Now look at the condition on n given in the question. Substitute the factorial formula you know. Try to expand the given term into simple parts. Simple part can be easily calculated. Now use the given expression to cancel or manipulate the things. Simplify the factorial only to an extent where you can use the remaining part for calculation.

Complete step-by-step solution -

Here you will just manipulate the term r and then use the concept of telescopic cancellation to cancel a lot of terms.
Factorial:- In mathematics, factorial is an operation, denoted by “(!)\left( ! \right) “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: n!=n×(n−1)×(n−2)...................×1n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1
For example: 5!=5×4×3×2×1=120.5!=5\times 4\times 3\times 2\times 1=120.
Note that we assume 0!=10!=1 . It is standard value. It has a wide range of applications in combinations.
Given summation in the question in terms of r, can be written as:
∑r=1nr(r!)\sum\limits_{r=1}^{n}{r\left( r! \right)}
By general algebra we know that r can be written as (r+1)−1\left( r+1 \right)-1 .
By substituting this into our summative, we can write it as:
∑r=1n((r+1)−1)(r!)\sum\limits_{r=1}^{n}{\left( \left( r+1 \right)-1 \right)\left( r! \right)}
By expanding the term inside the bracket, we get it as:
∑r=1n(r+1)r!−r!\sum\limits_{r=1}^{n}{\left( r+1 \right)r!-r!} .
By basic knowledge of factorials, we know that (r−1)r!\left( r-1 \right)r! can be written as (r+)!\left( r+ \right)! . By substituting this into our summation, we get the value of it as: ∑r=1n(r+1)!−r!\sum\limits_{r=1}^{n}{\left( r+1 \right)!-r!} ,
By expanding the summation, we can write it in the form of: ((n+1)!−n!)+(n!−(n−1)!)+.....................(3!−2!)+(2!−1!)\left( \left( n+1 \right)!-n! \right)+\left( n!-\left( n-1 \right)! \right)+.....................\left( 3!-2! \right)+\left( 2!-1! \right) .
By telescopic cancellation, we cancel a lot of terms and, we get: (n+1)!−1!=∑r=1nr(r!)\left( n+1 \right)!-1!=\sum\limits_{r=1}^{n}{r\left( r! \right)} .
So, the value of the summation is given by expression (n+1)!−1!\left( n+1 \right)!-1!.

Note: Be careful while writing r in different forms. Because it is the main point of solution, the idea is brought into action because we need summation of n terms. So, the best method is telescopic cancellation, to generate that we need a difference between two terms. To create that, we use this idea.