Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The sum of \[({1^2} - 1 + 1)(1!) + ({2^2} - 2 + 1)(2!) + ... + ({n^2} - n + 1)(n!)\] is
A. \[(n + 2)!\]
B. \[(n - 1)\left( {(n + 1)!} \right) + 1\]
C. \[(n + 2)! - 1\]
D. \[n\left( {(n + 1)!} \right) - 1\]

seo-qna
Last updated date: 17th Jun 2024
Total views: 372.9k
Views today: 10.72k
Answer
VerifiedVerified
372.9k+ views
Hint: Here in this question first we will write the given equation in the general form of a variable. The general form of an equation is the form in which when we put the value of the integers, we will get back our basic equation. Then we will simplify the equation to get the final answer.

Complete Complete Step by Step Solution:
The given equation is \[({1^2} - 1 + 1)(1!) + ({2^2} - 2 + 1)(2!) + ... + ({n^2} - n + 1)(n!)\]
Let us assume \[S = ({1^2} - 1 + 1)(1!) + ({2^2} - 2 + 1)(2!) + ... + ({n^2} - n + 1)(n!)\].
We can write it in the general form of a variable as \[S = \sum\limits_{r = 1}^n {({r^2} - r + 1)(r!)} \]
Simplifying above equation, we get
\[ \Rightarrow S = \sum\limits_{r = 1}^n {({r^2} - r + 1)(r!)} = \sum\limits_{r = 1}^n {({r^2} - 1 - r + 1 + 1)(r!)} \]
\[ \Rightarrow S = \sum\limits_{r = 1}^n {\left( {({r^2} - 1) - (r - 2)} \right)(r!)} = \sum\limits_{r = 1}^n {\left( {(r - 1)(r + 1) - (r - 2)} \right)(r!)} \]
Now multiplying \[r!\] to the terms we get, we get
\[ \Rightarrow S = \sum\limits_{r = 1}^n {\left( {(r - 1)(r + 1)r! - (r - 2)r!} \right)} = \sum\limits_{r = 1}^n {\left( {(r - 1)(r + 1)r! - (r - 2)r!} \right)} \]
Now expanding the above equation by putting the value of \[r\] as 1, 2, 3 to \[n\], we get
 \[ \Rightarrow S = \left[ {(1 - 1)(1 + 1)! - (1 - 2)1!} \right] + .... + \left[ {(n - 1)(n + 1)! - (n - 2)n!} \right]\]
Subtracting the terms, we get
\[ \Rightarrow S = \left[ {(0)(2)! - ( - 1)1!} \right] + \left[ {(1)(3)! - (0)2!} \right] + .... + \left[ {(n - 1)(n + 1)! - (n - 2)n!} \right]\]
Cancelling out the similar terms, we get
\[ \Rightarrow S = - ( - 1) + (n - 1)(n + 1)! = (n - 1)(n + 1)! + 1\]
Hence, \[(n - 1)(n + 1)! + 1\] is the sum of the given equation.

So, option B is the correct option.

Note:
Here, we need to know the way of writing an equation in the general form of a variable with its variable range. Generally, the value of the variable is positive natural numbers or integers.
Factorial of a number is equal to the multiplication or product of all the positive integers smaller than or equal to the number. In addition, factorial of 1 is always equals to 1 and factorial of 0 is equal to 1. The factorial of a number is always positive, it can never be negative and the factorial of a negative number is not defined.
Example of factorial:\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]