The sum of the squares of the first fifty integers

The sum of the squares of the first fifty integers

Some general formulae:

sum of squares of the first n positive or negative integers (1,2,3..., or -1 ,-2, -3,...)
=n(n+1)(2n+1)/6

sum of squares of the first n positive or negative odd integers (1,3,5..., or -1 ,-3, -5,...)
=n(4n^2-1)/3

sum of squares of the first n positive or negative even integers (1,3,5..., or -1 ,-3, -5,...)
=2n(n+1)(2n+1)/3


1) The sum of the squares of the first fifty positive integers is 42,925. What is the sum of the squares of the first fifty positive odd integers?
S1=50(4*50^2-1)/3=166650

2) The sum of the squares of the first fifty positive integers is 42,925. What is the sum of the squares of the first fifty negative even integers?
S2=2*50*(50+1)*(2*50+1)/3=171700

3) The sum of the squares of the first fifty positive even integers is 171,700. What is the sum of the squares of the first fifty positive even integers?
S3=2*50*(50+1)*(2*50+1)/3=171700


2008-01-06 22:42:12 補充：
Oops: correction required to the last general formulasum of squares of the first n positive or negative even integers (2,4,6... or -2, -4, -6...)=2n(n 1)(2n 1)/3

2008-01-07 07:53:36 補充：
If you are interested in how the formulae are derived, here is a link for the proof of the first formula(S(n)=n(n+1)(2n+1)/6). The use of mathematical induction is required.http://pirate.shu.edu/~wachsmut/ira/infinity/answers/sm_sq_cb.html

2008-01-07 07:59:55 補充：
Once we have established S1(n)=n(n+1)(2n+1)/6, the sum of the first n even integers is simply 4 times S(n), since S3(n)=2^2+4^2+6^2+...=4(1^2+2^2+3^2+...)=4*n(n+1)(2n+1)/6=2n(n+1)(2n+1)/3.

2008-01-07 08:02:16 補充：
The sum of the first n odd integers is simply the difference between the sum of the first 2n integers and the first n even integers, i.e.S2(n)=S1(2n)-S3(n)=2n(2n 1)(4n 1)/6 - 2n(n 1)(2n 1)/3= n(4n^2-1)/3as shown above.

2008-01-07 08:02:51 補充：
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