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Question: The value of \(1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + 10 \cdot 10! = .......\) A.\(11!\) B...

The value of 1⋅1!+2⋅2!+3⋅3!+...+10⋅10!=.......1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + 10 \cdot 10! = .......
A.11!11!
B.11!+111! + 1
C.11!−111! - 1
D.11⋅10!11 \cdot 10!

Explanation

Solution

Hint : Here, we are given the sum of the series with factorial numbers. We need to understand the concept of factorial for this. We will use the formula for the sum of the series for solving this question as we are given a sum of a series in which a number is multiplied by its own factorial starting from 1 to 10.

Complete step-by-step answer :
Here, we are asked to find the sum of the given series
S=1⋅1!+2⋅2!+3⋅3!+...+10⋅10!S = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + 10 \cdot 10!
We can also write this series in the form of summation as:
S=∑r=110r⋅r!S = \sum\limits_{r = 1}^{10} {r \cdot r!}
Here, rr can be rewritten as r+1−1r + 1 - 1.We will put this value in the main equation.

⇒S=∑r=110(r+1−1)⋅r! ⇒S=∑r=110((r+1)!−r!)  \Rightarrow S = \sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right) \cdot r!} \\\ \Rightarrow S = \sum\limits_{r = 1}^{10} {\left( {\left( {r + 1} \right)! - r!} \right)} \\\

Now, we will put the value of r as 1, 2, 3,…., 10 in the equation. Therefore we get,
⇒S=(2!−1!)+(3!−2!)+(4!−3!)+...+(11!−10!)\Rightarrow S = \left( {2! - 1!} \right) + \left( {3! - 2!} \right) + \left( {4! - 3!} \right) + ... + \left( {11! - 10!} \right)
Here, we can see that all the terms get cancelled out except −1! - 1!and 1!=11! = 1. Therefore, we can say that
⇒S=−1!+11!\Rightarrow S = - 1! + 11!
We know that 1!=11! = 1
⇒S=11!−1\Rightarrow S = 11! - 1
Thus, our answer is 1⋅1!+2⋅2!+3⋅3!+...+10⋅10!=11!−11 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + 10 \cdot 10! = 11! - 1
So, the correct answer is “Option C”.

Note : Here, we have determined the sum of given series from 1 to 10. We can aslo derive the general formula for this type of series from 1 to n.
For that we will take S=1⋅1!+2⋅2!+3⋅3!+...+n⋅n!S = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n!
We can also write this series in the form of summation as:
S=∑r=1nr⋅r!S = \sum\limits_{r = 1}^n {r \cdot r!}
Here, rr can be rewritten as r+1−1r + 1 - 1.We will put this value in the main equation.

⇒S=∑r=1n(r+1−1)⋅r! ⇒S=∑r=1n((r+1)!−r!)  \Rightarrow S = \sum\limits_{r = 1}^n {\left( {r + 1 - 1} \right) \cdot r!} \\\ \Rightarrow S = \sum\limits_{r = 1}^n {\left( {\left( {r + 1} \right)! - r!} \right)} \\\

Now, we will put the value of r as 1, 2, 3,…., n in the equation. Therefore we get,
⇒S=(2!−1!)+(3!−2!)+(4!−3!)+...+((n+1)!−n!)\Rightarrow S = \left( {2! - 1!} \right) + \left( {3! - 2!} \right) + \left( {4! - 3!} \right) + ... + \left( {\left( {n + 1} \right)! - n!} \right)
Here, we can see that all the terms get cancelled out except −1! - 1!and (n+1)!\left( {n + 1} \right)!. Therefore, we can say that
⇒S=−1!+(n+1)!\Rightarrow S = - 1! + \left( {n + 1} \right)!
We know that 1!=11! = 1
⇒S=(n+1)!−1\Rightarrow S = \left( {n + 1} \right)! - 1
Thus, we can say that 1⋅1!+2⋅2!+3⋅3!+...+n⋅n!=(n+1)!−11 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n! = \left( {n + 1} \right)! - 1 .
By using this general formula for the given series, we can directly find the answer for any given value of n.