I have no idea what those numbers say or mean.Probably because we haven't learn't about whatever you're talking about yet.

sorry but I haven't even learnt what an integer is.

when did you learn?

The differences between integers, rationals, and reals don't really have to be well defined (bad pun...) to understand them.  Just think of an integer as any whole number, without a decimal or fraction, a ration is anything that can be put in the form a/b (where a and b are integers), and real includes all the numbers without a complex part (i).

Last edited by Ricky (2006-03-02 05:02:27)

I'm expressing kk! of the form f(k+1)-f(k) so the sum

When I do this all f-s between 2 and n reduct as you see because f(k) thakes part in f(k+1)-f(k) with sign "-" and in f(k)-f(k-1) whit sign "+".
So there left only -f(1) and +f(n+1):

Now it's not hard to prove that kk!= (k+1)!-k!:
(k+1)!-k!= ((k+1)k!)-k!=k!(k+1-1)=kk!
so for every k kk!= (k+1)!-k!=f(k+1)-f(k),where f(x)=x!
So the sum:

,
which have to be proven.

Last edited by krassi_holmz (2006-03-02 16:24:50)

ganesh wrote:

MI # 1

Prove that the sum of n terms of
1*1! + 2*2! + 3.3! + 4*4!+.....n*(n+1)!
is (n+1)!-1.

You wrote n=1 instead of n

Let f(n)donate the series (|Last term is n*n!).

The statement is true for some p.