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Old 25th July 2012, 06:19 PM   #1 (permalink)
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Maths quiz

I was thinking of using summation... Am I on the right track?
.
Find the sum of
1x1! + 2x2! + 3x3! +...+ 9x9!
Without using calculator.

Edit:I got the answer already


use summation
Method of different
nn! = (n+1)! - n!

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Last edited by .Memo; 25th July 2012 at 08:33 PM.
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Old 25th July 2012, 10:33 PM   #2 (permalink)
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Re: Maths quiz

What a waste of thread lol, allow me to further extend the problem.

Evaluate, without using a calculator,
[1!(1˛+1)+2!(2˛+1)+3!(3˛+1)+...+9!(9˛+1)]÷10!

Source: Myself!!! I spent about 20min making this problem, don't let it go to waste. Good luck solving!
Inspired by: Your maths quiz lols

Edit: In case you doubt me, Computer's answer

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Last edited by Betakuwe; 25th July 2012 at 10:59 PM.
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Old 27th July 2012, 07:58 PM   #3 (permalink)
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Re: Maths quiz

Originally Posted by Betakuwe View Post
What a waste of thread lol, allow me to further extend the problem.

Evaluate, without using a calculator,
[1!(1˛+1)+2!(2˛+1)+3!(3˛+1)+...+9!(9˛+1)]÷10!

Source: Myself!!! I spent about 20min making this problem, don't let it go to waste. Good luck solving!
Inspired by: Your maths quiz lols

Edit: In case you doubt me, Computer's answer
Neat modification to the original problem!

Note that n!(n^2+1)=(n!*n)*n+n!=n*(n+1)!-n*n!+n!=n*(n+1)!-(n-1)*n!
also see (n-1)![(n-1)^2+1]=(n-1)*(n)!-(n-2)*(n-1)!
So we see a telescope sum(method of difference)
1!(1˛+1)+2!(2˛+1)+3!(3˛+1)+...+9!(9˛+1)]=9*(10!) - 0 = 9*(10!)
So the answer is then 9.

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Old 27th July 2012, 08:08 PM   #4 (permalink)
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Re: Maths quiz

Originally Posted by .Memo View Post
I was thinking of using summation... Am I on the right track?
.
Find the sum of
1x1! + 2x2! + 3x3! +...+ 9x9!
Without using calculator.

Edit:I got the answer already


use summation
Method of different
nn! = (n+1)! - n!
In all honesty, Cambridge will guide you in your approach if they give you the question like this. Just saying.

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Old 27th July 2012, 08:40 PM   #5 (permalink)
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Re: Maths quiz

Originally Posted by Tetrahedron View Post
In all honesty, Cambridge will guide you in your approach if they give you the question like this. Just saying.
It's not a exam question, but rather a small bi-monthly quiz set by my school.

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Old 27th July 2012, 10:25 PM   #6 (permalink)
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Re: Maths quiz

Originally Posted by .Memo View Post
It's not a exam question, but rather a small bi-monthly quiz set by my school.
Hahas, they get these questions from past Singapore Mathematics Olympiads.

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Old 27th July 2012, 10:55 PM   #7 (permalink)
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Re: Maths quiz

Originally Posted by megaman123 View Post
Neat modification to the original problem!

Note that n!(n^2+1)=(n!*n)*n+n!=n*(n+1)!-n*n!+n!=n*(n+1)!-(n-1)*n!
also see (n-1)![(n-1)^2+1]=(n-1)*(n)!-(n-2)*(n-1)!
So we see a telescope sum(method of difference)
1!(1˛+1)+2!(2˛+1)+3!(3˛+1)+...+9!(9˛+1)]=9*(10!) - 0 = 9*(10!)
So the answer is then 9.
There are quite a few approaches to this problem. My original solution when I made this was very long, but the simplest way I found to go about doing this is the following:
Note that
n!(n˛+1)
=n!((n+1)˛-2n)
=n!(n+1)˛ - 2(n!*n)
=(n+1)*(n+1)! - 2(n!*n)
=(n+2)!-(n+1)! - 2[(n+1)!-n!] ——(Using n*n!=(n+1)! - n!)
Then by telescoping, we easily get the answer 9.

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