Calculus 1, Session 30 -- Riemann sums

[jana]

We will try to derive the formula for the sum of squares and cubes in this comment. Note that ∑_(k=1)^n (3k^2 + 3k + 1) = ∑_(k=1)^n [(k + 1)^3 - k^3] = (n + 1)^3 - 1^3 = n^3 + 3n^2 + 3n. Note that: ∑_(k=1)^n (3k^2 + 3k + 1) = ∑ 3k^2 + ∑ 3k + ∑ 1 (all from k = 1 to n). Note that this is just equal to 3(∑k^2) + 3∑k + n = n^3 + 3n^2 + 3n. Therefore, using simple algebra we have ∑_(k=1)^n k^2 = 1/3 (n^3 + 3n^2 + 2n - 3∑_(k=1)^n k) = n(n+1)(2n+1)/6.

I will provide a proof of the sum of cubes as well, but to clear up notation, assume that ∑ represents ∑_(k=1)^n.

Note that (k - 1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1. Thus, for all k in Z we have ∑ (4k^3 + 6k^2 + 4k + 1) = ∑[(k + 1)^4 - k^4] = (n + 1)^4 - 1 = n^4 + 4n^3 + 6n^2 + 4n. Note that ∑(4k^3 + 6k^2 + 4k + 1) = 4∑k^3 + 6∑k^2 + 4∑k + ∑1. Thus, we have

∑(4k^3 + 6k^2 + 4k + 1) = 4∑k^3 + 6∑k^2 + 4∑k + ∑1 = 4∑k^3 + 2n^3 + 5n^2 + 4n after simplifying each individual sum. Therefore, and finally, from earlier we had that 4∑k^3 + 2n^3 + 5n^2 + 4n = n^4 + 4n^3 + 6n^2 + 4n, so getting the sum of cubes alone, we see that ∑k^3 = (n(n + 1)/2)^2. Notice that this is just the formula for the sum of the first n positive integers squared.

Hope I didn't make any mistakes since looking at this was pretty difficult.

[jana]

43:58 I wouldn't agree that these sums are calc 2. I learned how to derive the third one using polynomial series in my algebra 2 class. The others we derived in my algebra 1 class.

[codyprochaska]

Great video. Shouldn't this be titled session 30 though? 29 was anti derivatives.