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Question: If \({a_n} = n\left( {n!} \right)\), then \(\sum\limits_{r = 1}^{100} {{a_r}} \) is equal to (A) ...

If an=n(n!){a_n} = n\left( {n!} \right), then ∑r=1100ar\sum\limits_{r = 1}^{100} {{a_r}} is equal to
(A) 101!101!
(B) 100!−1100! - 1
(C) 101!−1101! - 1
(D) 101!+1101! + 1

Explanation

Solution

Transform the term an=n(n!){a_n} = n\left( {n!} \right) as an=n(n!)=(n+1)!−n!{a_n} = n\left( {n!} \right) = \left( {n + 1} \right)! - n! . Now use this expression to calculate values at n=1,2,3.....,r−1,rn = 1,2,3.....,r - 1,r . Add all the equations together and find a simpler expression for series a1+a2+a3+......+ar−1+ar{a_1} + {a_2} + {a_3} + ...... + {a_{r - 1}} + {a_r}. For finding the required summation put r=100r = 100 and find the correct option from the question.

Complete step by step solution:
Here in the given problem, the value of an{a_n} is defined as n(n!)n\left( {n!} \right) which is the product of the number ′n′'n' and its factorial. With this information, we need to find the value of ∑r=1100ar\sum\limits_{r = 1}^{100} {{a_r}}
Factorial of a number is the product of a natural number and all the natural number below it. For any number, mm the factorial is denoted by m!m!
The expression n(n!)n\left( {n!} \right) can be represented as:
⇒n×n!=(n+1−1)×n!=(n+1)×n!−n!\Rightarrow n \times n! = \left( {n + 1 - 1} \right) \times n! = \left( {n + 1} \right) \times n! - n!
As we know the property of factorial, i.e. m!=m×(m−1)!m! = m \times \left( {m - 1} \right)! . Using this property, we get:
an=n(n!)=(n+1)!−n!{a_n} = n\left( {n!} \right) = \left( {n + 1} \right)! - n!
Now, using the above relation, we can form a pattern for different values of the number ′n′'n'
For n=1n = 1 ; a1=(1+1)!−1!=2!−1!{a_1} = \left( {1 + 1} \right)! - 1! = 2! - 1!
For n=2n = 2 ; a2=(2+1)!−2!=3!−2!{a_2} = \left( {2 + 1} \right)! - 2! = 3! - 2!
For n=3n = 3 ; a3=(3+1)!−3!=4!−3!{a_3} = \left( {3 + 1} \right)! - 3! = 4! - 3!
⋮\vdots
For n=rn = r ; ar=(r+1)!−r!{a_r} = \left( {r + 1} \right)! - r!
On adding all the values of a1,a2,a3......ar{a_1},{a_2},{a_3}......{a_r} we get:
⇒a1+a2+a3+......+ar−1+ar=(2!−1!)+(3!−2!)+(4!−3!)+......+(r!−(r−1)!)+((r+1)!−r!)\Rightarrow {a_1} + {a_2} + {a_3} + ...... + {a_{r - 1}} + {a_r} = \left( {2! - 1!} \right) + \left( {3! - 2!} \right) + \left( {4! - 3!} \right) + ...... + \left( {r! - \left( {r - 1} \right)!} \right) + \left( {\left( {r + 1} \right)! - r!} \right)
Now we can see that all the terms between first and the last will get reduced to zero but addition with the same number with the opposite sign. The remaining term will determine the value of summation:
⇒a1+a2+a3+......+ar−1+ar=(r+1)!−1!=(r+1)!−1\Rightarrow {a_1} + {a_2} + {a_3} + ...... + {a_{r - 1}} + {a_r} = \left( {r + 1} \right)! - 1! = \left( {r + 1} \right)! - 1
Therefore, for the value of summation ∑r=1100ar\sum\limits_{r = 1}^{100} {{a_r}} will represent:
⇒∑r=1100ar=a1+a2+a3+.....+a99+a100\Rightarrow \sum\limits_{r = 1}^{100} {{a_r}} = {a_1} + {a_2} + {a_3} + ..... + {a_{99}} + {a_{100}}
We can use the above relation of ar{a_r} to find the above-required value:
⇒∑r=1100ar=a1+a2+a3+.....+a99+a100=(100+1)!−1=101!−1\Rightarrow \sum\limits_{r = 1}^{100} {{a_r}} = {a_1} + {a_2} + {a_3} + ..... + {a_{99}} + {a_{100}} = \left( {100 + 1} \right)! - 1 = 101! - 1
Therefore, we get the required summation of ∑r=1100ar\sum\limits_{r = 1}^{100} {{a_r}} as 101!−1101! - 1

Hence, the option (C) is the correct answer.

Note:
In questions like this try to manipulate the expression to get a pattern in the summation. This makes it easier to calculate the long series. The symbol ∑r=1100ar\sum\limits_{r = 1}^{100} {{a_r}} is known as sigma and represents the summation of terms with initial values as r=1r = 1 to the final value as r=100r = 100.