Question

Prove by mathematical induction 1+3+5+...+(2n-1)=n^2

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. 

Answer 2

Answer:

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

2

be the given statement

Step 1: Put n = 1

Then, L.H.S = 1

R.H.S = (1)

2

= 1

∴. L.H.S = R.H.S.

⇒ P(n) istrue for n = 1

Step 2: Assume that P(n) istrue for n = k.

∴ 1 + 3 + 5 + ..... + (2k - 1) = k

2

Adding 2k + 1 on both sides, we get

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

2

+ (2k + 1) = (k + 1)

2

∴ 1 + 3 + 5 + ..... + (2k -1) + (2(k + 1) - 1) = (k + 1)

2

⇒ P(n) is true for n = k + 1.

∴ by the principle of mathematical induction P(n) is true for all natural numbers 'n'

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

2

, for all n ϵ n