1!+2!+3!+4!+.....+n!=?.... How do I find the answer of this sum?
#Sum of Factorials of 1st n natural numbers.
idle rose
craggy shellBOT
wind elm
answer is S = 1 - 2^5 + 2(4!) (1 - 5/2 + 1/(2*5))
= 1 - 32 + 48 (1 - 2.5 + 0.1)
= 153
idle rose
answer is S = 1 - 2^5 + 2(4!) (1 - 5/2 + 1/(2*5))
= 1 - 32 + 48 (1 - 2.5 + 0.1)
...
How?
wind elm
How?
Evaluate 2^5 = 32.
Evaluate 4! = 4 x 3 x 2 x 1 = 24.
Evaluate the expression inside the parentheses: 1 - 5/2 + 1/(2*5) = 1 - 2.5 + 0.1 = -1.4.
Evaluate 2(4!) = 2 x 24 = 48.
Substitute the values from steps 1, 3, and 4 into the original expression: S = 1 - 32 + (48 x -1.4)
= 1 - 32 - 67.2
= -98.2, Finally, add the negative sign to get the absolute value of S:
|S| = 98.2.
idle rose
wait, absolute value of S can never be negative cause each term is positive... And the sum of n terms in that series cannot be a finite value... The answer must be in terms of n
grave night
