Using Proof By Induction I am trying to prove the following:
$n^2 = \sum_{i=1} ^{n} (2i-1) $ for all $n\geq 1$
Here is my solutions so Far:
Base Case: $n=1, LHS: 2(1)-1 = 1, RHS = 1^2 = 1, True$
Induction Hypothesis:
Assume true for $n=k$
$k^2 =\sum_{i=1} ^{k} (2i-1) $ for some $k\geq 1$
Induction Step: Should be True for $n=k+1$
$(k+1)^2 =\sum_{i=1} ^{k+1} (2i-1) $ for some $k\geq 1$
However here is where I get stuck
We want to show that, given the inductive hypothesis, $$(k+1)^2 =\sum_{i=1} ^{k+1} (2i-1)$$
LHS: $k^2 + 2k+1$.
The RHS
$\begin{align} \sum_{i=1} ^{k+1} (2i-1)& \overset{IH}{=} k^2 + 2(k+1) - 1 \\ & = k^2 + 2k +2-1 \\ \\ & = k^2 + 2k + 1\end{align}$
Hence, given the inductive hypothesis, it follows that $(k+1)^2 =\sum_{i=1} ^{k+1} (2i-1)$. Hence, $$n^2 = \sum_{i=1}^{n} (2i-1)$$