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 $n$
 A: The sequence is
$$1=1 \  , \ 3+5 = 8 \ , \ 7+9+11 = 27 \ , \ \dots$$
So you can see that
$$a_n = \sum_{k=1}^{1+2+...+n} (2k-1) - \sum_{k=1}^{n-1} a_k =$$
$$\sum_{k=0}^{n(n+1)/2} (2k-1) - a_{n-1} =\frac{1}{4}n^2(n+1)^2 - \sum_{k=1}^{n-1} a_k$$
because you can view the $n$-th term as the sum of $1+2+\dots+n$ odd numbers minus the sum of all the other previous terms ($27 = 1+3+5+7+9+11 - (1)-(3+5)$).
So the base step is clear, now for the inductive one we use strong induction (we say it works from $1$ to $n-1$) then
$$a_{n} = \frac{1}{4}n^2(n+1)^2 - \sum_{k=1}^{n-1} k^3 = \frac{1}{4}n^2(n+1)^2 -\frac{1}{4}(n-1)^2n^2 = n^3$$
so the formula is proven.
A: 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.
A: Let us find the general term of $1,3,7,13,\cdots$
As the next consecutive differences are $2,4,6,\cdots$
$$T_n=an(n-1)+b(n-1)(n-2)+cn(n-2)$$
$n=1\implies1=c(1-2),c =-1$
$n=2\implies3=2a$
$n=3,7=6a+2b+3c=3(3)+2b+3(-1),2b=1$
$$\implies T_n=\cdots=n^2-n+1$$
We need  the sum of $n$ terms from $k=n^2-n+1$ to $(n+1)^2-(n+1)+1-1$ with the common difference $=2$
which is $$n(n^2-n+1)+\sum_{k=1}^n(2k)=n^3-n^2+n+\dfrac n2(2+(n-1)2)=?$$
