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# Advanced Math/Solve a simpler problem and look for a pattern

dwilson wrote at 2006-07-09 20:39:44
Question

Find the sum of two hundred even numbers from 0 to 398.

A minor edit to the original answer:

S = 200/2 [2(0) + (200 - 1)2]

= 100 [0 + (199)2]

= 100 (398)

= 39800

dwilson wrote at 2006-07-09 20:56:42
Question

Find the sum of two hundred even numbers from 0 to 398.

2006.07.09

Looking for a pattern.

Label the numbers in the series

0, 2, 4, 6, 8, ..., 394, 396, 398.

1, 2, 3, 4, 5, ..., 198, 199, 200.

Successive sums

up to numbers labelled 1, 2, 3, 4, 5

are 0, 2, 6, 12, 20

These sums can be factored into

1*0, 2*1, 3*2, 4*3, 5*4.

So sum to number labelled 200, ie 398 is

200*199 = 39800.

Chantz wrote at 2008-11-12 05:03:04
Is there an error in your calculation?  It seems that the section (200-1) would result in 199, not 198.  Therefore, when the 198 is multiplied by 2 and then the results multiplied by 100, the answer would be 39600.  But since (200-1) is actually 199, the equation would be as follows:

S=200/2[2(0)+(200-1)2]

S=100[0+(199)2]

S=100(398)

S=39,800

karen wrote at 2012-11-01 23:50:45
In the equation above, isn't 200-1=199 NOT 198.  Therefore the answer would be (199)2=398.  398 X 100 = 39,800

Ahmed Salami wrote at 2012-11-09 10:53:28
Yes, you've all been right in pointing this out. Thank you.

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