Question

Use the principle of mathematical induction to prove that 1 +3 +5 + … +r = n2 for all

n>0 where r is an odd integer & n is the number of terms in the sum.​

Answer 1

Given:

P(n) : 1 + 3 + 5 + … + r = n²

Where n > 0 and r is an odd integer

To find:

Prove that 1 + 3 + 5 + … + r = n²

Solution:

Since, r is an odd integer

So, we can write r as (2n - 1)

∴ P(n) : 1 + 3 + 5 + … + (2n - 1) = n²

Step 1: Check P(1) is true or not

Put n = 1

LHS = 1

RHS = 1² = 1

∴ LHS = RHS = 1

Step 2: Let P(k) is true

Put n = k

So, 1 + 3 + 5 + … + (2k - 1) = k²

Step 3: Prove that P(k+1) is also true

LHS = 1 + 3 + 5 + … + (2k - 1) + (2k + 1)

Since, 1 + 3 + 5 + … + (2k - 1) = k²

∴ k² + (2k + 1)

k² + 2k + 1

(k + 1)²                                                                   Note: a² + b² + 2ab = (a + b)²

So, LHS = (k + 1)²    

RHS = (k + 1)²

∴ LHS = RHS = (k + 1)²

Therefore, P(n) is true for n = k + 1

Therefore, by the principle of mathematical induction P(n) if true for all the natural numbers.