Proof by induction: $\sum^{2n-1}_{i=1} (2i-1)=(2n-1)^2$

Question

$\sum^{2n-1}_{i=1} (2i-1)=(2n-1)^2$

I get stuck after proving the base case is true. Usually with induction I assume the left and right sides are equal at some k, but I'm not sure how to approach this problem since the left side is a sum.

Answer 1

If $\sum_{i=1}^{2n-1}(2i-1)=(2n-1)^2$, then\begin{align}\sum_{i=1}^{2(n+1)-1}(2i-1)&=\sum_{i=1}^{2n+1}(2i-1)\\&=\left(\sum_{i=1}^{2n-1}(2i-1)\right)+(4n-1)+(4n+1)\\&=(2n-1)^2+8n\\&=4n^2+4n+1\\&=(2n+1)^2.\end{align}

Answer 2

You can notice that, when you replace $n$ with $n+1$, the sum contains the sum for $n$, with some further terms; for instance $$ \sum_{i=1}^{n+1}i = \biggl(\,\sum_{i=1}^n i\biggr)+(n+1) $$ In your case, $$ \sum_{i=1}^{2(n+1)-1} (2i-1) = \sum_{i=1}^{2n+1} (2i-1) = \biggl(\,\sum_{i=1}^{2n-1} (2i-1)\biggr) +(2\cdot2n-1)+(2\cdot(2n+1)-1) $$ because the last summation misses the last two terms, the one for $i=2n$ and the one for $i=2n+1$.

Now you can replace the summation by $(2n-1)^2$ because of the induction hypothesis and try to see whether this produces the same as $(2(n+1)-1)^2$.