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Question: The value of \[\dfrac{1}{2}\]! \[ + \]\[\dfrac{2}{3}\]! \[ + \].........\[ + \]\[\dfrac{{999}}{{1000...

The value of 12\dfrac{1}{2}! + $$$$\dfrac{2}{3}! ++......... + $$$$\dfrac{{999}}{{1000}}! is

Explanation

Solution

In this question, we have to find the value of a given series based on factorial. We will proceed by writing the numerator of each terms of the given series as the difference of its denominator minus one( i.e. 12!=(2−1)2!\dfrac{1}{{2!}} = \dfrac{{(2 - 1)}}{{2!}} ). Now we will distribute the denominator over numerator
(i.e. (2−1)2!=22!−12!\dfrac{{(2 - 1)}}{{2!}} = \dfrac{2}{{2!}} - \dfrac{1}{{2!}} ) and then cancel the same terms to get the resultant solution of the series.

Complete answer:
This question is based on sequence and series of factors. A factorial in mathematics is a function that multiplies a number by every number below. it is denoted by the symbol ! . For example , 5!=5×4×3×2×15! = 5 \times 4 \times 3 \times 2 \times 1 , 6!=6×5×4×2×16! = 6 \times 5 \times 4 \times 2 \times 1 and so on.
Consider the given question,
We have to find the value of 12\dfrac{1}{2} factorial + $$$$\dfrac{2}{3} factorial ++......... + $$$$\dfrac{{999}}{{1000}} factorial
Let , x=12!+23!+..............+9991000!x = \dfrac{1}{{2!}} + \dfrac{2}{{3!}} + .............. + \dfrac{{999}}{{1000!}} , where symbol !! denote factorial in mathematics .
The above expression can also be written as ,
⇒x=(2−1)2!+(3−1)3!+..............+(1000−1)1000!\Rightarrow x = \dfrac{{(2 - 1)}}{{2!}} + \dfrac{{(3 - 1)}}{{3!}} + .............. + \dfrac{{(1000 - 1)}}{{1000!}}
distributing the denominator over numerator , we have
⇒x=22!−12!+33!−13!+..............+10001000!−11000!\Rightarrow x = \dfrac{2}{{2!}} - \dfrac{1}{{2!}} + \dfrac{3}{{3!}} - \dfrac{1}{{3!}} + .............. + \dfrac{{1000}}{{1000!}} - \dfrac{1}{{1000!}}
Cancelling the common term from numerator and denominator we have,
⇒x=11!−12!+12!−13!+..............+1999!−11000!\Rightarrow x = \dfrac{1}{{1!}} - \dfrac{1}{{2!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + .............. + \dfrac{1}{{999!}} - \dfrac{1}{{1000!}}
Now we can see that in the above series, each term gets cancelled except the first and last term. Hence, we have
⇒x=11!−11000!\Rightarrow x = \dfrac{1}{{1!}} - \dfrac{1}{{1000!}}
On simplifying, we get
⇒x=1000!−11000!\Rightarrow x = \dfrac{{1000! - 1}}{{1000!}}
Hence, the value of the given series 12\dfrac{1}{2} factorial + $$$$\dfrac{2}{3}factorial ++......... + $$$$\dfrac{{999}}{{1000}} factorial is 1000!−11000!\dfrac{{1000! - 1}}{{1000!}}.

Note:
Factorial is a function that multiplies a number by every number less than itself. For example: n!=n(n−1)(n−2)(n−3).......(3)(2)(1)n! = n(n - 1)(n - 2)(n - 3).......(3)(2)(1) where nn is any integer greater than 11.
We define zero factorial equal to one. (i.e. 0!=10! = 1 ).
Factorial of Negative integers are not defined.