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Calculus II
Some Series used with Riemann sum



1. Sum of the n first odd integers


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

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

Adding the two equalities yields:

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

Therefore

2Sn = 2 n2. That is Sn = n2
    n        
Sn  =   Σ   (2i - 1)   =   n2
    i = 1  

2. Sum of the n first integers


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

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

Adding the two equalities yield:

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

Therefore

Sn = n (n + 1) /2
    n        
Sn  =   Σ   i   =   n(n + 1)/2
    i = 1  

3. Sum of the n squered first integers


Sn2 = 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 Sn2 + 3 Sn + n =
3 Sn2 + 3 n(n + 1)/2 + n

Therefore

3 Sn2 + 3 n(n + 1)/2 + n = (n + 1)3 - 1 =
n3 + 3 n2 + 3 n + 1 - 1 = n3 + 3 n2 + 3 n
3 Sn2 = n3 + 3 n2 + 3 n - 3 n(n + 1)/2 - n =
n3 + 3 n2 + 2 n - (3/2) n2 - (3/2) n =

Then

6Sn2 = 2 n3 + 3 n2 + n = n(n + 1)(2n + 1)

or

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




  


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