Arithmetic series

Question

Series of odd number is 1+3+5+7+...

 

 a)Prove that the sum of the first n odd integers is n squared 

 

 b) Check answer to a) by finding S1. S2. S3 and S4

Answer 1

1, 3, 5, 7, ... is an arithmetic sequence with common difference 2.  

 

a_n, the nth term of the sequence is given by the formula a_n = a₁+(n-1)d

 

So, for this sequence, a_n = 1 + (n-1)(2) = 2n-1

 

The sum of the first n terms of an arithmetic sequence is given by the formula:

 

 S_n= (n/2)[a₁+a_n]

 

So, for the sequence 1,3,5,7,...,

 

S_n = 1+3+5+7+...+(2n-1) = (n/2)[1 + (2n-1)]

 

                                     = (n/2)(2n) = n²

 

Answer 2

Proof by mathematical induction:

an odd number can by written as  2n-1 where n=1, 2, 3, ...

prove that 1+3+5+7+...+2n-1=n²

show that this is true for k=1, assume this is true for some odd integer 2k-1, and finally show this is true for 2(k+1)-1

if k=1 then we have 1 integer, the number 1, 2*1-1=1², 2-1=1², 1=1², 1=1

assume this is true for 2k+1

show this is true for 2(k+1)-1

1+3+5+7+9+...+(2k-1)+[2(k+1)-1]=

(1+3+5+7+9+...+(2k-1))+[2(k+1)-1]=

(1+3+5+7+9+...+(2k-1)=k² is assumed to be true

k²+[2(k+1)-1]=

k²+[2k+2-1]=

k²+2k+1=

(k+1)²

1+3+5+7+9+...+2k-1+[2(k+1)-1]=(k+1)²

S(1)=1

S(2)=1+3=4=2²

S(3)=1+3+5=9=3²

S(4)=1+3+5+7=16=4²