Prove the following by the principle of mathematical induction: `1+3+5++(2n-1)=n^2i.e.` the sum of first `n` odd natural numbers is `n^2dot`

Explanation

Suppose P(n) denotes the statement.
1+3+5+……to n terms =`n^(2)`.
i.e. 1+3+5+….(2n-1)=`n^(2)` ….(i)
for n=1 (I) means `1=1^(2) i.e. 1=1` which is true
`therefore` P(1) is true.
Let us assume that P(k) is true, `k in N`
i.e. 1+3+5+....(2k-1)=`k^(2)` ....(ii)
Now, 1+3+5+......+(k+1) terms
=1+3+5+.....+[2(k+1)-1]
=1+3+5+....(2k+1)
=[1+3+5+....+(2k-1)]+(2k+1)
=`k^(2)+2k+1` [by using (ii)]
`=(k+1)^(2)`
`Rightarrow` P(n) is true for n=k+1
Therefore, by principle of mathematical, induction, P(n) is true for all n.
Hence, the sum of first n odd natural number is `n^(2)`.