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## #1 2006-02-28 15:48:15

ganesh
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### Mathematical Induction

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.

Character is who you are when no one is looking.

## #2 2006-02-28 17:11:19

krassi_holmz
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### Re: Mathematical Induction

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 (2006-02-28 17:14:10)

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## #3 2006-03-02 03:39:10

espeon
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### Re: Mathematical Induction

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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## #4 2006-03-02 03:42:20

krassi_holmz
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### Re: Mathematical Induction

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 (2006-03-02 04:00:39)

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## #5 2006-03-02 03:55:50

espeon
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### Re: Mathematical Induction

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

Presenting the Prinny dance.
Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

## #6 2006-03-02 04:00:53

krassi_holmz
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### Re: Mathematical Induction

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

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## #7 2006-03-02 04:03:04

espeon
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### Re: Mathematical Induction

when did you learn?

Presenting the Prinny dance.
Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

## #8 2006-03-02 04:09:12

krassi_holmz
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### Re: Mathematical Induction

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

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## #9 2006-03-02 05:01:55

Ricky
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### Re: Mathematical Induction

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)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #10 2006-03-02 15:07:14

ganesh
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### Re: Mathematical Induction

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.

## #11 2006-03-02 16:21:32

krassi_holmz
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### Re: Mathematical Induction

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)

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## #12 2006-09-23 12:48:14

mapu
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### Re: Mathematical Induction

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)

## #13 2011-03-01 18:38:07

G-man
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### Re: Mathematical Induction

#### 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!......