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#1 2006-02-27 16:48:15

ganesh
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Registered: 2005-06-28
Posts: 14,879

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.


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#2 2006-02-27 18:11:19

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

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 (2006-02-27 18:14:10)


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#3 2006-03-01 04:39:10

espeon
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Registered: 2006-02-05
Posts: 2,586

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 2006-03-01 04:42:20

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

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 (2006-03-01 05:00:39)


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#5 2006-03-01 04:55:50

espeon
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Registered: 2006-02-05
Posts: 2,586

Re: Mathematical Induction

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


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#6 2006-03-01 05:00:53

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

Re: Mathematical Induction

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


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#7 2006-03-01 05:03:04

espeon
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Registered: 2006-02-05
Posts: 2,586

Re: Mathematical Induction

when did you learn?


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#8 2006-03-01 05:09:12

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

Re: Mathematical Induction

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


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#9 2006-03-01 06:01:55

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

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


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#10 2006-03-01 16:07:14

ganesh
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Registered: 2005-06-28
Posts: 14,879

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.


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#11 2006-03-01 17:21:32

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

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


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#12 2006-09-22 14:48:14

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

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)

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#13 2011-02-28 19:38:07

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

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.





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