These sums were taken from the book Number Theory by George E. Andrews
What is the equation for each of the following sums that gives the sum for a given n:
(1) 1+2+3+⋯+n=n(n+1)2 since:
- 2(1+2+3+⋯+n)= ([1+2+3+⋯+n]+[n+(n−1)+(n−2)+⋯+1])= (1+n)+(2+[n−1])+(3+[n−2])+⋯+(n+1)=(n)(n+1)
(2) 12+22+32+⋯+n2=n(n+1)(2n+1)6 since:
- Binomial Theorem: (x+y)^p = \sum\limits_{k=0}^{p}{p \choose k}x^{p-k}{y^{k}}
- (n-1)^3 = n^3 -3n^2 +3n - 1
- n^3 - (n-1)^3 = 3n^2 - 3n + 1
- n(n+1)(2n+1) = (2n+1)(n^2 + n) = 2n^3 + 2n^2 + n^2 + n = 2n^3 + 3n^2 + n
- n^3 = n^3 - (n-1)^3 + (n-1)^3 - (n-2)^3 + \dots + 1^3 - 0 = 3(n^2 + (n-1)^2 + \dots 1^2) - 3(n + (n-1) + \dots + 1) + n(1) = 3(1^2 + 2^2 + \dots + n^2) - 3(\dfrac{n(n+1)}{2}) + n
- n^3 + 3\left(\dfrac{n(n+1)}{2}\right) - n = \dfrac{2n^3 + 3n^2 + 3n - 2n}{2} = \dfrac{2n^3 + 3n^2 + n}{2} = \dfrac{(n)(n+1)(2n+1)}{2} = 3(n^2 + (n-1)^2 + \dots 1^2)
(3) 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2 since:
- (n-1)^4 = n^4 - 4n^3 + 6n^2 - 4n + 1
- n^4 - (n-1)^4 = 4n^3 - 6n^2 + 4n - 1
- [n(n+1)]^2 = n^2(n^2 + 2n + 1) = n^4 + 2n^3 + n^2
- n^4 = n^4 - (n-1)^4 + (n-1)^4 - (n-2)^4 + \dots + 1^4 - 0^4 = 4(n^3 + (n-1)^3 + \dots + 1^3) - 6(n^2 + (n-1)^2 + \dots + 1^2) + 4(n + (n-1) + \dots + 1) - n = 4(1^3 + 2^3 + \dots + n^3) - 6\left(\dfrac{2n^3 + 3n^2 +n}{6}\right) + 4\left(\dfrac{n(n+1)}{2}\right) -n
- n^4 + 6\left(\dfrac{2n^3 + 3n^2 +n}{6}\right) - 4\left(\dfrac{n^2 + n)}{2}\right) + n = n^4 + 2n^3 + 3n^2 + n - 2n^2 -2n +n = n^4 + 2n^3 + n^2 = [n(n+1)]^2 = 4(1^3 + 2^3 + \dots + n^3)
- x(1 + x + x^2 + x^3 + \dots + x^n) = x + x^2 + x^3 + x^4 + \dots + x^{n+1}
- (x-1)(1 + x + x^2 + x^3 + \dots + x^n) = x^{n+1} - 1
- n(n+1) = n^2 + n
- 1\times2 + 2\times3 + 3\times4 + \dots + (n)(n+1) = (1^2 + 1) + (2^2 + 2) + (3^2 + 3) + \dots + + (n^2 + n) = (1 + 2 + 3 + \dots + n) + (1^2 + 2^2 + 3^2 + \dots + n^2) = \dfrac{n(n+1)}{2} + \dfrac{n(n+1)(2n+1)}{6} = \dfrac{(3n + 2n^2 +n)(n+1)}{6} = \dfrac{(n^2+2n)(n+1)}{3}
- 1 + 3 + 5 + \dots + (2n-1) = 2(1 + 2 + 3 + \dots + n) - n = n(n+1) - n = n^2 + n -n
- \dfrac{1}{1\times2} + \dfrac{1}{2\times3} + \dfrac{1}{3\times4} + \dots + \dfrac{1}{n(n+1)} = \dfrac{2-1}{1\times2} + \dfrac{3-2}{2\times3} + \dfrac{4-3}{3\times4} + \dots + \dfrac{(n+1)-n}{n(n+1)} = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + \dots + \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right) = 1 - \dfrac{1}{n+1} = \dfrac{n+1 - 1}{n+1}