As given by the question
There are given that the series
[tex]1+3+5+\cdots+(2n-1)=n^2[/tex]Now,
For step 1:
Put n=1
Then LHS =1
And
[tex]\begin{gathered} R\mathrm{}H\mathrm{}S=(n)^2 \\ =(1)^2 \\ =1 \end{gathered}[/tex]So,
[tex]\therefore L.H.S=R.H.S[/tex]P(n) is true for n=1.
Now,
Step 2:
Assume that P(n) istrue for n=k
Then,
[tex]1+3+5+\cdots+(2n-1)=k^2[/tex]Adding 2k+1 on both sides
So, we get:
[tex]1+3+5\ldots+(2k-1)+(2k+1)=k^2+(2k+1)=(k+1)^2[/tex]P(n) is true for n=k+1
By the principle of mathematical induction P(n) is true for all natural numbers n.
Hence,
[tex]1+3+5+\cdots+(2n-1)=n^2[/tex]For all n.
Hence proved.
