The double factorial, denoted by n!!, returns the product of all of the odd integers that are less than or equal to n. For example, 7!! = 7 times 5 times 3 times 1. What is the units digit of 1!! + 3!! + 5!! + 7!! + ... + 49!!?

MathCuber
Nov 19, 2018

#2**+1 **

Here are two different sums done on two different computers and they both agree!!

∑[(2n + 1)!!, n, 0, 24]=**59654360733417426078066607239049**

a=listfor(n, 0, 24,(2*n+1)!!;sum(a)=**59 6543607334 1742607806 6607239049**

**Rom: You have a mistake somewhere!! Here is another way: (49 - 5) / 2 + 1= 23 x 5=115+3!!+1!! =119.**

**Also: ****∑[(2n+1)!!, n, 0, 24] mod 10 = 9!**

Guest Nov 19, 2018

edited by
Guest
Nov 19, 2018

edited by Guest Nov 19, 2018

edited by Guest Nov 19, 2018

#3**+1 **

As Rom points out....5!! + 7!! + 9!! + .....+ 49!! will end in 5

[ Each term will have 5 * an odd product ...i.e., ending in a "5"....and we will have 23 terms....so....an odd number of summed terms ending in "5" will also end in "5" ]

And

1!! = 1

3!! = 3

So....the sum ends in 1 + 3 + 5 = 9

CPhill
Nov 19, 2018