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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 40

Hi all,

I have following string, and need to develop a single formula for all terms.

n, (n-1), (n-2)!, (n-3)!+(n-4)!+(n-5)!+...(n-(n-1))!, ((n-4)!+(n-5)!+(n-6)!+...(n-(n-1))!)+((n-5)!+(n-6)!+(n-7)!+...(n-(n-1))!)+...+(n-(n-1))!, ((n-5)!+(n-6)!+(n-7)!+...(n-(n-1))!)+((n-6)!+(n-7)!+(n-8)!+...(n-(n-1))!)+...+(n-(n-1))!

as illustrated after 3rd term, each sub-term (n-m)! in next term turn to (n-(m-1))!+ (n-(m-2))!+ (n-(m-3))!+....+(n-(n-1))!

any idea is appreciated

thank you:)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,940

Hi kappa_am;

A single formula? I am not following you. Can you explain a little better?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 40

Hi,

lets put first two terms aside.

Doing so, I have a string of terms starts with (n-2)!. in this string each term is constructed according to the sub-terms of the previous term for example:

fist term is (n-2)! = (n-2)+(n-3)+(n-4)+(n-5)+...(n-(n-1)); the second term is (n-3)!+(n-4)!+(n-5)!+....+(n-(n-1))!

the third term will be summation of second side of the following equations:

in second term we have (n-3)! which is (n-3)+(n-4)+(n-5)...+(n-(n-1)) so one sub-term in third term is (n-4)!+(n-5)!+(n-6)!+...(n-(n-1))!

in second term we have (n-4)! which is (n-4)+(n-5)+(n-6)...+(n-(n-1)) so another sub-term in third term is (n-5)!+(n-6)!+(n-7)!+...(n-(n-1))!

...

in second term we have (n-(n-2)) which is (n-(n-2))+(n-(n-1)) so the last sub-term in third term is (n-(n-1))!

the other terms in this string are constructed as it is stated according to the previous term. I need a single formula that enables me to obtain each term.

an algorithm or a program in C language is also appreciated.

Thank you very much

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,376

hi kappa_am

An exclamation mark (factorial) means the terms are multiplied.

eg 6! = 6 x 5 x 4 x 3 x 2 x 1

In post 3 you have put + not x.

fist term is (n-2)! = (n-2)+(n-3)+(n-4)+(n-5)+...(n-(n-1));

Which do you want?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 40

Sorry,

I mean summation, not multiplication.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,376

It would help if you re-write the question.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 40

I have a string of terms starts with ∑(n-m) {down limit of summation is m=2 and upper limit is m=(n-1). in this string each term is constructed according to the sub-terms of the previous term for example:

fist term is ∑(n-2) = (n-2)+(n-3)+(n-4)+(n-5)+...(n-(n-1)); the second term is ∑(n-3)+∑(n-4)!+∑(n-5)+....+∑(n-(n-1))

the third term will be summation of second side of the following equations:

in second term we have ∑(n-3) which is (n-3)+(n-4)+(n-5)...+(n-(n-1)) so one sub-term in third term is ∑(n-4)+∑(n-5)+∑(n-6)+...∑(n-(n-1))

in second term we have ∑(n-4) which is (n-4)+(n-5)+(n-6)...+(n-(n-1)) so another sub-term in third term is ∑(n-5)+∑(n-6)+∑(n-7)+...∑(n-(n-1))

...

in second term we have (n-(n-2)) which is (n-(n-2))+(n-(n-1)) so the last sub-term in third term is ∑(n-(n-1))=1

the other terms in this string are constructed as it is stated according to the previous term. In above equations "∑" means summation of the term located front of it starting from indicated number to n-1. for example ∑(n-4)= (n-4)+(n-5)+(n-6)+....+(n-(n-1).

I need a single formula that enables me to obtain each term. an algorithm or a program in C language is also appreciated.

Thank you very much

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