Calculus II
Some Series used with Riemann sum

1. Sum of the n first odd integers

S_n = 1 + 3 + 5 + ... + 2n - 1 (n terms)

S_n = 2n - 1 + ... + 3 + 2 + 1

Adding the two equalities yields:

2 S_n = (2n - 1 + 1) + (2n - 3 + 3) + ... + (2n - 3 + 3) + (2n - 1 + 1) =
(2n) + (2n) + ... + (2n) + (2n) = n x (2n) = 2 n²

Therefore

2S_n = 2 n². That is S_n = n²

		n
Sn	=	Σ		(2i - 1)	=	n2
		i = 1

2. Sum of the n first integers

S_n = 1 + 2 + 3 + ... + n - 2 + n - 1 + n

S_n = n + n -1 + n - 2 + ... + 3 + 2 + 1

Adding the two equalities yield:

2 S_n = (n + 1) + (n + 1) + (n + 1) + ... + (n + 1) = n(n + 1)

Therefore

S_n = n (n + 1) /2

		n
Sn	=	Σ		i	=	n(n + 1)/2
		i = 1

3. Sum of the n squered first integers

S_n² = 1² + 2² + 3² + ... +
(n - 2)² + (n - 1)² + n² .

We have :

(n + 1)³ = n³ + 3 n² + 3 n + 1
2³ - 1³ = 3 (1)² + 3 (1) + 1
3³ - 2³ = 3 (2)² + 3 (2) + 1
4³ - 3³ = 3 (3)² + 3 (3) + 1
... = ...
(n + 1)³ - n³ = 3 n² + 3 n + 1

Adding all the equalities yield:

(n + 1)³ - 1³ = 3 (1² + 2² + 3² + ... +n²) + 3 (1 + 2 + 3 + ... + n) + 1 + 1 + 1 + ... + 1) =
3 S_n² + 3 S_n + n =
3 S_n² + 3 n(n + 1)/2 + n

Therefore

3 S_n² + 3 n(n + 1)/2 + n = (n + 1)³ - 1 =
n³ + 3 n² + 3 n + 1 - 1 = n³ + 3 n² + 3 n
3 S_n² = n³ + 3 n² + 3 n - 3 n(n + 1)/2 - n =
n³ + 3 n² + 2 n - (3/2) n² - (3/2) n =

Then

6S_n² = 2 n³ + 3 n² + n = n(n + 1)(2n + 1)

or

S_n² = n(n + 1)(2n + 1)/6

		n
Sn2	=	Σ		i2	=	n(n + 1)(2n + 1)/6
		i = 1