Answer
384.6k+ 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:\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
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:\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)