first part (a) i m taking 4 odd term 1,3,5,7 . sum of 4 odd term=16 which is same to conclude (no of odd terms)^2=sum of consecutive n odd terms
same as induction given here,
$\sum$(for i=0, (2i+1)=1 , for i=1,(2i+1)=3 , for i=2,(2i+1)=5, for i=3,(2i+1)=7)=16 (i.e nothing but (no of terms)^2=(4)^2=16)
(b)since in question they have said to use result from above
i m taking above equation
$\sum_{0}^{n-1}(2i+1)= n^{2}$ ,now breaking it into two parts [1 (for i=0 , (2i+1) gives 1 )]+[$\sum_{1}^{n}(2i+1)$] (here if u carefully see then we have increased lower as well as upper limit by 1)=n^2+2n+1(we added 2n+1 in rhs side because upper limit has increased by 1 so nth term will give us 2n+1 which is added in rhs side)
$1+\sum_{i=1}^{n}(2i+1)=n^{2}+2n+1$
$\sum_{i=1}^{n}(2i+1)=n^{2}+2n$
$\sum_{i=1}^{n}(2i)+\sum_{1}^{n}1=n^{2}+2n$
$\sum_{i=1}^{n}(2i)+n=n^{2}+2n$
$\sum_{i=1}^{n}(2i)=n^{2}+n$
$2\sum_{i=1}^{n}(i)=n^{2}+n$
$\sum_{i=1}^{n}(i)=(n^{2}+n)/2$