Original post by B.Carmine
Can someone help me here
prrof that Sum[ r(r!) ] = (n+1)! - 1
So far i have (k+1)! - 1 + (k+1)(k+1)!
pls help
prrof that Sum[ r(r!) ] = (n+1)! - 1
So far i have (k+1)! - 1 + (k+1)(k+1)!
pls help
factorise the first and last terms.
Original post by B.Carmine
Can someone help me here
prrof that Sum[ r(r!) ] = (n+1)! - 1
So far i have (k+1)! - 1 + (k+1)(k+1)!
pls help
prrof that Sum[ r(r!) ] = (n+1)! - 1
So far i have (k+1)! - 1 + (k+1)(k+1)!
pls help
Try taking out a common factor of (k+1)! from (k+1)!+(k+1)(k+1)!.
Original post by the bear
factorize that mofo
what is a mofo?
Original post by notnek
Try taking out a common factor of (k+1)! from (k+1)!+(k+1)(k+1)!.
But surely i can't with that cheeky -1?
Original post by TeeEm
what is a mofo?
it is an expression with two or more factorials.
Original post by the bear
it is an expression with two or more factorials.
never heard of it ...
Original post by B.Carmine
Ok, so, i'll do it anyway.
(k+!)![k+2 - 1/(k+1)!]
= (k+2)(k+1)! - 1
....... not right, right?
(k+!)![k+2 - 1/(k+1)!]
= (k+2)(k+1)! - 1
....... not right, right?
What is (k+2)(k+1)! ?
Original post by 16Characters....
What is (k+2)(k+1)! ?
oh
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