
If \[S = 1\left( 1 \right)! + 2\left( 2 \right)! + 3\left( 3 \right)! + .... + n\left( n \right)!\] , then which of the following statement is true?
A. \[\dfrac{{S + 1}}{{n!}} \in \] integer
B. \[\dfrac{{S + 1}}{{n!}} \notin \] integer
C. \[\dfrac{{S + 1}}{{n!}}\] cannot be discussed
D. None of these
Answer
164.4k+ views
Hint: Here, the equation of the sum of the factorial terms is given. First, simplify the given sum by using the various properties of the factorial. After that, solve the right-hand side and get the sum in the minimum number of terms. In the end, divide the sum by \[n!\] and check the division with the given options to get the required answer.
Formula Used: \[n! = n\left( {n - 1} \right)\left( {n - 2} \right)....3 \times 2 \times 1\]
\[n! = n\left( {n - 1} \right)!\]
Complete step by step solution: The given equation of the sum is \[S = 1\left( 1 \right)! + 2\left( 2 \right)! + 3\left( 3 \right)! + .... + n\left( n \right)!\].
Let’s simplify the given sum.
\[S = \left( {2 - 1} \right)\left( 1 \right)! + \left( {3 - 1} \right)\left( 2 \right)! + \left( {4 - 1} \right)\left( 3 \right)! + .... + \left( {n + 1 - 1} \right)\left( n \right)!\]
Solve the right-hand side.
\[S = \left( {2 \times 1!} \right) - 1! + \left( {3 \times 2!} \right) - 2! + \left( {4 \times 3!} \right) - 3! + .... + \left( {\left( {n + 1} \right) \times n!} \right) - n!\]
Now apply the formula of the factorial \[n! = n\left( {n - 1} \right)!\].
\[S = 2! - 1! + 3! - 2! + 4! - 3! + .... + \left( {n + 1} \right)! - n!\]
\[ \Rightarrow S = - 1! + \left( {n + 1} \right)!\]
\[ \Rightarrow S = \left( {n + 1} \right)! - 1\]
\[ \Rightarrow S + 1 = \left( {n + 1} \right)!\]
Now, divide both sides by \[n!\].
\[\dfrac{{S + 1}}{{n!}} = \dfrac{{\left( {n + 1} \right)!}}{{n!}}\]
Simplify the right-hand side by applying the formula \[n! = n\left( {n - 1} \right)!\].
\[\dfrac{{S + 1}}{{n!}} = \dfrac{{\left( {n + 1} \right)n!}}{{n!}}\]
\[ \Rightarrow \dfrac{{S + 1}}{{n!}} = \left( {n + 1} \right)\]
Since \[n\] is an integer, \[\left( {n + 1} \right)\] is also an integer.
Option ‘A’ is correct
Note: The factorial of a number is a product of all whole numbers less than that number up to 1.
It is also used to calculate the number of possible ways in which a selected number of objects can be arranged among themselves.
Formula Used: \[n! = n\left( {n - 1} \right)\left( {n - 2} \right)....3 \times 2 \times 1\]
\[n! = n\left( {n - 1} \right)!\]
Complete step by step solution: The given equation of the sum is \[S = 1\left( 1 \right)! + 2\left( 2 \right)! + 3\left( 3 \right)! + .... + n\left( n \right)!\].
Let’s simplify the given sum.
\[S = \left( {2 - 1} \right)\left( 1 \right)! + \left( {3 - 1} \right)\left( 2 \right)! + \left( {4 - 1} \right)\left( 3 \right)! + .... + \left( {n + 1 - 1} \right)\left( n \right)!\]
Solve the right-hand side.
\[S = \left( {2 \times 1!} \right) - 1! + \left( {3 \times 2!} \right) - 2! + \left( {4 \times 3!} \right) - 3! + .... + \left( {\left( {n + 1} \right) \times n!} \right) - n!\]
Now apply the formula of the factorial \[n! = n\left( {n - 1} \right)!\].
\[S = 2! - 1! + 3! - 2! + 4! - 3! + .... + \left( {n + 1} \right)! - n!\]
\[ \Rightarrow S = - 1! + \left( {n + 1} \right)!\]
\[ \Rightarrow S = \left( {n + 1} \right)! - 1\]
\[ \Rightarrow S + 1 = \left( {n + 1} \right)!\]
Now, divide both sides by \[n!\].
\[\dfrac{{S + 1}}{{n!}} = \dfrac{{\left( {n + 1} \right)!}}{{n!}}\]
Simplify the right-hand side by applying the formula \[n! = n\left( {n - 1} \right)!\].
\[\dfrac{{S + 1}}{{n!}} = \dfrac{{\left( {n + 1} \right)n!}}{{n!}}\]
\[ \Rightarrow \dfrac{{S + 1}}{{n!}} = \left( {n + 1} \right)\]
Since \[n\] is an integer, \[\left( {n + 1} \right)\] is also an integer.
Option ‘A’ is correct
Note: The factorial of a number is a product of all whole numbers less than that number up to 1.
It is also used to calculate the number of possible ways in which a selected number of objects can be arranged among themselves.
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