Riemenn sum, quick question.?

Question

Please see this layman's pic, since trying to write it here is a hellish ordeal: http://img98.imageshack.us/img98/4360/reimennrl7.jpg

The first part is the actual problem. The second part, why is the problem split into two sums? Is it because the k below sigma in the first problem is = 3, and we have to set it to 1? I don't see the relationship, thanks!

Answer 1

Simplified directly:

summation of k^2 from 3 to 10
= 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2
= 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100
= 380

Simplified indirectly:

= summation of K^2 from 1 to 10 - summation of k^2 from 1 to 2

= (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2)
- (1^2 + 2^2)
= 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2
= 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100
= 380

Most math problems have some intended lesson.

If we know the the sum of the squares of the first n natural numbers is (1 / 6)n(n + 1)(2n + 1), then summations of k^2 from 1 to n can be calculated quickly.

summation of k^2 from 3 to 10

= summation of k^2 from 1 to 10 - summation of k^2 from 1 to 2

= (1 / 6)10(10 + 1)(2(10) + 1) - (1 / 6)2(2 + 1)(2(2) + 1)
=(1 / 6)(10)(11)(21) - (1 / 6)(2)(3)(5)
= 385 - 5
= 380

Answer 2

That's simply saying that
3^2 + 4^2 + ... + 10^2 =
(1^2 + 2^2 + 3^2 + 4^2 + ... + 10^2) - (1^2 + 2^2)

∑ from 3 to 10 = ∑ from 1 to 10 - ∑ 1 to 2

The reason the problem is split like that is because the sum of all the squares from 1 to N^2 is a known formula:
N(N + 1)(2N + 1) / 6
So the sum of squares from 3 to 10 can be found by applying this formula to N = 2 and N = 10
10*(10+1)*(2*10+1)/6 - 2*(2+1)*(2*2+1)/6
= 385 - 5 = 380

Answer 3

First of all, this is not a "Riemann sum". It's just an ordinary summation expression. In this case, it's about adding the first 10 squares, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. The first summation is adding all the squares from 9 to 100, which can be expressed as the sum of all the squares from 1 to 100 MINUS the sum of all the squares from 1 to 4. Get it now?

Answer 4

By splitting the problem into 2 sums, you can use a known formula for the sums and subtract.

n
∑x^2 = (1/6)(n + 3n^2 + 2n^3) = n(n + 1)(2n + 1)/6
x=1

so
10
∑x^2 = (10*11*21 - 2*3*5)/6 = 380
3

For small n, it's easier to do the manual calculation than to derive the formula, but for larger n the formula greatly simplifies calculation.