Find a pattern and prove it by mathematical induction: $1 = 1$
$3 + 5 = 8$
$7 + 9 + 11 = 27$
$13 + 15 + 17 + 19 = 64$
Etc...

I am having trouble seeing a pattern with this, I know it is relatable with Fibonacci Numbers but I am having trouble grasping this topic
 A: The first term it's $n^2-n+1$ 
and we have a sum of $n$ terms of an arithmetic progression with $d=2$.
Thus, $$(n^2-n+1)+(n^2-n+1+2)+...+(n^2-n+1+2(n-1))=$$
$$=n(n^2-n+1)+2\cdot\frac{n(n-1)}{2}=n^3.$$
A: The pattern is 
\begin{align}
& (n^2-n+1) + (n^2-n+3) + \cdots + (n^2+n-3)+(n^2+n-1)\\
& =\frac{n}{2}(n^2-n+1+n^2+n-1)\\
& = n^3.
\end{align}
This is the sum of $n$ terms of an arithmetic progression.
A: We can continue the above pattern as follows:
$$ 21 + 23 + 25 + 27 + 29 = 125 = 5^3.$$
$$ 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3. $$
We note that the first odd natural number is $1$, which we can write as $$ 1 = 2(1) - 1;$$
the second odd natural number is $3$, which can be written as 
$$ 3 = 2(2) - 1; $$
the third odd natural number is $5$, which can be written as 
$$ 5 = 2(3) - 1; $$
and so on and so forth. 
We also note that, for each $n \in \mathbb{N}$, at step $n$, we add all the odd natural numbers from the ($ 1 + \sum_{k=1}^n k $)th to the ($\sum_{k=1}^{n+1} k$)th to obtain $n^3$.
Now 
$$ 1 + \sum_{k=1}^n k  = 1 + \frac{n(n+1)}{2} = \frac{n^2 + n + 2}{2}, $$
and 
$$ \sum_{k=1}^{n+1} k = \frac{ (n+1)(n+2) }{2} = \frac{ n^2 + 3n + 2 }{2}. $$
So we can write
$$ \sum_{k= \frac{n^2 + n + 2}{2}}^{\frac{ n^2 + 3n + 2 }{2} } (2k-1) = n^3 $$
for every $n \in \mathbb{N}$. 
This can be rewritten as 
$$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k \ - \  \left[ \frac{ n^2 + 3n + 2 }{2}  - \frac{n^2 + n + 2}{2} \right] = n^3 $$
for each $n \in \mathbb{N}$. 
Or, 
$$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k \ - \ n = n^3 $$
for each $n \in \mathbb{N}$. 
Or, 
$$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k = n^3 + n $$
for each $n \in \mathbb{N}$. 
We can verify the last identity using mathematical induction. 
Hope this works out for you now.
A: Note that the first term in each sum $1,3,7,13,21,\cdots$ are quadratic (consider their finite differences and the differences of the differences) thus a formula for this seuence is $n(n-1)+1$. Each sum has $n$ terms and each summand increases by $2$, thus
\begin{eqnarray*}
\sum_{i=1}^{n}(n(n-1)+2i-1)=n^2(n-1)+n(n+1)-n= \color{red}{n^3}.
\end{eqnarray*}
So each sum is $n^3$ as observed by lulu in the comments.
A: Hint: $$ [1+2\cdot( 1+ 2+\ldots +(n-1))]\cdot n + 2(1+2+\ldots +(n-1)) $$
You can then use Gauss formula for $1+ 2+\ldots +(n-1)$.
As in $n=4$:
$$13 + (13 + 2) + (13 +4) + (13 +6) = 13\cdot 4 + 2\cdot(1+2+3)$$
and 
$$ 13 = 1 + 2\cdot( 1+  2 + 3)$$
A: Let $a_n$ the $n$-th term of the sequence.
The sum of the $n$ first terms is the sum of the first $\frac{n(n+1)}{2}$ odd integers, which is $\left( \frac{n(n+1)}{2} \right)^2$.
So that:
\begin{equation}
a_1+\ldots +a_n = \left( \frac{n(n+1)}{2} \right)^2
\end{equation}
\begin{equation}
a_1+\ldots +a_{n-1} = \left( \frac{(n-1)n}{2} \right)^2
\end{equation}
Then
$$a_n = \left( \frac{n(n+1)}{2} \right)^2 - \left( \frac{(n-1)n}{2} \right)^2 = \\
\left( \frac{n(n+1)}{2}+\frac{(n-1)n}{2} \right)\left( \frac{n(n+1)}{2}-\frac{(n-1)n}{2} \right) = 
\left( \frac{n^2+n}{2}+\frac{n^2-n}{2} \right)\left( \frac{n^2+n}{2}-\frac{n^2-n}{2} \right) = 
n^2 \cdot n = n^3
$$
