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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,614

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.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

k.k!= (k+1)k! - k! = (k+1)! - k!

so

1.1!+2.2!+3.3!+4.4!+...+(n).(n)!**/*not n+1*/**=2!-1!+3!-2!+4!-3!+...+(n+1)!-n!= (n+1)!-1

*Last edited by krassi_holmz (2006-02-27 18:14:10)*

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**espeon****Real Member**- Registered: 2006-02-05
- Posts: 2,586

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

It's never early to learn something:

k! means the product of all positive integers, less or equal to k:

k!=1.2.3. ... .k

*Last edited by krassi_holmz (2006-03-01 05:00:39)*

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**espeon****Real Member**- Registered: 2006-02-05
- Posts: 2,586

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

Presenting the Prinny dance.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

OK.

I'll leave you to your maths teacher to teach you what's an integer.

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**espeon****Real Member**- Registered: 2006-02-05
- Posts: 2,586

when did you learn?

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I'm from Bulgaia. I can't tell you.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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 (2006-03-01 06:02:27)*

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,614

krassi_holmz, I am not able to understand your proof. If no other member posts the solution, I shall do it before posting the next problem.

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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 (2006-03-01 17:24:50)*

IPBLE: Increasing Performance By Lowering Expectations.

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**mapu****Member**- Registered: 2006-09-22
- Posts: 1

CAn you help me with a question about mathematical induction?

Prove that

k^4= (k^5)/5 + (k^4)/2 +(K^3)/3 -k/30

by mathematical induction for the next term (k+1)

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**G-man****Member**- Registered: 2011-02-28
- Posts: 16

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.

Maths!......

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