Question

Prove by principle of mathematical induction 1+3+5+....+(2n-1)=n^2 for all n belonging to n

Answer 1

Proof by induction on n:

 

Step 1:  prove that the equation is valid when n = 1

 

              When n = 1, we have (2(1) - 1) = 12, so  the statement                        holds for n = 1.

 

Step 2:  Assume that the equation is true for n, and prove that the                   equation is true for n + 1.

 

              Assume:  1 + 3 + 5 + ... + (2n - 1) = n2

 

             Prove:  1 + 3 + 5 +...+ (2(n + 1) - 1) = (n + 1)2

 

                 Proof:  1 + 3 + 5 +... + (2(n + 1) - 1) 

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

                              = n2 + (2n + 2 - 1)  (by assumption)

                              = n2 + 2n + 1

                              = (n + 1)2

 

So, by induction, for every positive integer n,

1 + 3 + 5 + ... + (2n - 1) = n2. 

 

 

1/21/2015 | Mark M.

 

Comment



Saroj N. | Prompt services for mathematics

0

the given series is in AP

with first term, a = 1 & common difference, d = 2

Let us assume that  Tn = 1+(n-1)*2 = 2n-1

 

Hence the (2n-1) is the nth term of AP.

 

Sum of AP Series  = n(a+l)/2  where l = Last term

                         = n (1+2n-1)/2 = 2n2/2= n2 (proved)