<s>[INST] Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$
 [/INST] Approaching by pattern,

*

*We know that sum of first $k$ odd numbers is $k^2$.

*We have $1,2,3,4\ldots$ terms which may be expressed as difference of triangular number of terms
$$ 1 = \underset{1}{\underbrace{1}} - \underset{0}{\underbrace{0}} $$ $$ 3+5 = \underset{\text{first 3 odd}}{\underbrace{(1+3+5)}} - \underset{\text{first odd}}{\underbrace{1}}$$ $$ 7+9+11 = \underset{\text{first 6 odd}}{\underbrace{(1+3+5+7+9+11)}}-\underset{\text{first 3 odd}}{\underbrace{(1+3+5)}}$$
In particular, $n^{th}$ sum is $$\text{sum of first $T_{n}$ odd numbers - sum of first $T_{n-1}$ odd numbers}$$

*

*Thus $n^{th}$ sum is $$ 1 = 1^2 - 0^2 $$ $$ 3+5 = 3^2 - 1^2$$ $$ 7+9+11 = 6^2-3^2$$ $$ 13+15+17+19 = 10^2-6^2$$
In particular, $n^{th}$ sum is $$T_{n}^2-T_{n-1}^2$$
$$=\dfrac{n^2(n+1)^2}{4}-\dfrac{n^2(n-1)^2}{4}$$
$$=\dfrac{n^2}{4}(4n)$$
$$=n^3$$

Note : Once one makes these observations by naked eye, one can simply finish the problem by writing down step $3$ directly.
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