# Arithmetic problem

### Arithmetic problem

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
miteshshanbhag

Posts: 3
Joined: Sun Sep 12, 2010 2:01 am
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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.

martosss

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Joined: Fri Jun 06, 2008 5:16 am
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ya...the 5th term is 30..not 32....sorry typing mistake...
the method is correct....great......
THANX A TON BUDDY.....

miteshshanbhag

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

miteshshanbhag

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### Re: arithmetic problem

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

leesajohnson

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