Prove $(2n + 1) + (2n + 3) + \cdots + (4n - 1) = 3n^2$ by induction This might be an easy problem for you, but I am having difficulties in understanding the formula.
As we can see, we have a pattern
$$2n + \text{odd number}$$
in
$$(2n + 1) + (2n + 3) + \cdots + (4n - 1),$$
except for the last term:
$$(4n - 1).$$
So it's difficult to use induction (at least for me).
If not for the last term, I would use the formula to sum all odd terms up to the $n$.
Plus, I could retrieve the sum of all $2n$ terms by multiplying by $n$.
With some other manipulations, I could probably prove this, but in this case, with that last term, I am not seeing what I can do, because of my lack of imagination. 
 A: $4n-1$ is the same as $2n+(2n-1)$, that is $2n+$ an odd number.
So it's the sum of all the $n$ odd numbers between $2n+1$ and $4n-1$, inclusive.
A: You wanted it by induction so here it is: For $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : (2n+1)+(2n+3)+\cdots+(4n-1)=3n^2.
$$
It may be easier to see what is going on if we rewrite $S(n)$ as
$$
S(n) : (2n+1)+(2n+3)+\cdots+(2n+2n-3)+(2n+2n-1)=3n^2.
$$
Base step ($n=1$): The statement $S(1)$ says $2\cdot 1+1=4\cdot 1-3$, and this is true.
Induction step: Fix some $k\geq 1$, and assume that
$$
S(k) : (2k+1)+(2k+3)+\cdots+(2k+2k-3)+(2k+2k-1)=3k^2
$$
is true. To be shown is that
$$
S(k+1) : \underbrace{(2(k+1)+1)+(2(k+1)+3)+\cdots+(2(k+1)+2(k+1)-1)}_{\text{LHS}}=\underbrace{3(k+1)^2}_{\text{RHS}}.
$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\text{LHS} &= (2(k+1)+1)+\cdots+(2(k+1)+2(k+1)-3)+(2(k+1)+2(k+1)-1)\\[0.5em]
&= (2k+1)+2+\cdots+(2k+2k-1)+2+(2k+2k+1)+2\\[0.5em]
&= (2k+1)+(2k+3)+\cdots+(2k+2k-1)+2k+(4k+3)\\[0.5em]
&= 3k^2+6k+3\qquad\text{(by inductive hypothesis)}\\[0.5em]
&= 3(k^2+2k+1)\\[0.5em]
&= 3(k+1)^2\\[0.5em]
&= \text{RHS},
\end{align}
we see that the right-hand side of $S(k+1)$ follows, completing the inductive step.
By mathematical induction, for all $n\geq 1, S(n)$ is true. $\blacksquare$
A: Hint: It might help to try to rewrite the terms in the same form as the first one, such as $2n+3=2(n+1)+1,$ $2n+5=2(n+2)+1,$ and so on, up to $4n-1=2(2n-1)+1.$
By rewriting in this way, we can rewrite the sum in terms of more familiar sums, for which we know the closed form.
Another (arguably simpler) form would be $2n+3=2n+1+2(1),$ $2n+5=2n+1+2(2),$ and so on, up to $4n-1=2n+1+2(n-1).$
A: As Henning Makholm's comment, we have the sum of all the $n$ odd numbers between $2n+1$ and $4n−1$, inclusive. Thus, you can see it as the sum of all numbers from $2n$ to $4n$ minus the sum of all even numbers in the same range:
\begin{align}(2n+1)+(2n+3)\cdots(2n+(2n-1))&=\sum_{i=2n}^{4n}i - \sum_{i=n}^{2n}2i \\\\ &=\left(\sum_{i=1}^{4n}i-\sum_{i=1}^{2n-1}i\right)-2\left(\sum_{i=1}^{2n}i-\sum_{i=1}^{n-1}i\right) \\ \\&=\left(\frac{(4n)(4n+1)}{2}-\frac{(2n-1)(2n)}{2}\right)\\ \\ &-2\left(\frac{(2n)(2n+1)}{2}-\frac{(n-1)(n)}{2}\right) \\ \\ &= 3n^2
\end{align}
The interesting thing is that we didn't use induction "directly", only the sum of the first $n$ numbers.
