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#1 2006-02-27 16:48:15
- Jai 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.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#2 2006-02-27 18:11:19
- 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 (2006-02-27 18:14:10)
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#3 2006-03-01 04:39:10
- 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 2006-03-01 04:42:20
- 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 (2006-03-01 05:00:39)
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#5 2006-03-01 04:55:50
- espeon
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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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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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Re: Mathematical Induction
when did you learn?
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#8 2006-03-01 05:09:12
- krassi_holmz
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Re: Mathematical Induction
I'm from Bulgaia. I can't tell you.
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#9 2006-03-01 06:01:55
- 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 (2006-03-01 06:02:27)
"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..."
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#10 2006-03-01 16:07:14
- Jai 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.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#11 2006-03-01 17:21:32
- 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 (2006-03-01 17:24:50)
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#12 2006-09-22 14:48:14
- 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)
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#13 2011-02-28 19:38:07
- G-man
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Re: 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.
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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