<s>[INST] 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
 [/INST] 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.
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