Arithmetic problem

Arithmetic problem

Postby miteshshanbhag » Sun Sep 12, 2010 2:05 am

pls tell me a formula to compute
0+2+6+12+20+32+...........+9900
9900 is the 100th term
the nos. are increasing in the order 2,4,6,8,10,......
i think it is a different type of arithmetic progression
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Postby martosss » Sun Sep 19, 2010 9:49 am

If they are increasing in this order, then the 5-th member should be 30 instead of 32!
Then try representing the set as
1+4+9+16+25...9801=1²+2²+...99²
1+2+3+4 +5+...99
____________________
2+6+12+20+30...9900

You have two new sets:
1+2+3+...99
and
1²+2²+3²+...99²
and you need to find their sum.
Now you need to find the formula for computing the value of each of the new sets, which is not that hard. ;)
the first one is n*(n+1)/2
The second I leave to you. :P

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Postby miteshshanbhag » Mon Sep 20, 2010 6:52 am

ya...the 5th term is 30..not 32....sorry typing mistake...
the method is correct....great...... :D
THANX A TON BUDDY..... :)

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Postby miteshshanbhag » Mon Sep 20, 2010 6:54 am

THE SECOND FORMULA IS
n*(n+1)*(2n+1)/6

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Re: arithmetic problem

Postby leesajohnson » Sat May 07, 2016 5:37 am

n*(n+1)*(2n+1)/6

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