Originally Posted by
Jameson Let P(n) be the statement that $\displaystyle P(n)=\sum_{i=n}^{2n-1}2i+1=3n^2$, for all non-negative n. Assume that P(n) is true for some $\displaystyle n \le k$
$\displaystyle P(k+1)=\sum_{i=k+1}^{2(k+1)-1}2i+1$
The upper limit of the sum is also 2k+1. Think about how this second sum compares to the first.
$\displaystyle \sum_{i=k+1}^{2(k+1)-1}2i+1=\left(\sum_{i=k}^{2k-1}2i+1\right)-(2k+1)+(4k+1)+(4k+2+1)$
I'll let you figure out why P(k+1) can be written like that. What extra terms does it contain that P(k) does not?