MHB What is the Sum of Factorials for a Specific Equation?

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Factorials Sum
AI Thread Summary
The discussion focuses on evaluating the sum of factorials for the equation $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$. It simplifies the expression using the identity $$\dfrac{n^2-2}{n!}=\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!}$$. The final evaluation leads to the result of $$3-\dfrac {1}{2011!}-\dfrac{2}{2012!}$$. This approach effectively breaks down the sum into manageable parts, highlighting the relationship between factorials and the series. The conclusion provides a clear answer to the original question posed in the thread.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Evaluate $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$
 
Last edited:
Mathematics news on Phys.org
anemone said:
Evaluate $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$
$\dfrac{n^2-2}{n!}=\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!}$

$\therefore \sum_{2}^{2012}(\dfrac{n^2-2}{n!})=$$ \sum_{2}^{2012}(\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!})$

$=3-\dfrac {1}{2011!}-\dfrac{2}{2012!}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top