There exists a sequence $\{x_n\}$ members of which satisfy the following equation:
$$x_{m+n} = x_m + x_n + m\cdot n,\; \forall m, n \in \mathbb N, \; x_1 = a $$ Find the general term of the sequence.
To solve this I started from the case when $m = n = 1$. That gives:
$$ x_2 = x_1 + x_1 + = 2x_1 +1 $$
I want to look at $x_3, x_4, ..., x_n$ and find some pattern. So for $m = 1$ and $n = 2$ we obtain:
$$ x_3 = x_1 + x_2 + 2 = x_1 + 2x_1 + 1 = 3x_1 + 3 $$
Continuing that process we obtain:
$$ x_1 = \color{green}1x_1 + \color{blue}0 \\ x_2 = \color{green}2x_1 + \color{blue}1 \\ x_3 = \color{green}3x_1 + \color{blue}3 \\ x_4 = \color{green}4x_1 + \color{blue}8 \\ x_5 = \color{green}5x_1 + \color{blue}{10} \\ x_6 = \color{green}6x_1 + \color{blue}{15} $$
And so on. The green values form a sequence of natural numbers. For the blue ones i referred to OEIS and the sequence is in the form ${1 \over 2} (n-1)$. Based on that the general term is:
$$ x_n = n(a + {1\over 2}(n-1)) $$
I've also looked at a similar sequence which satisfies the following:
$$ x_{m+n} = x_m + x_n + m + n,\; \forall m, n \in \mathbb N $$
And again I was able to search for the sequence in OEIS but could not deduce it myself without a lookup. Also it is not going to always work since one may imagine sequences which match for the first $n$ terms and starting from $n+1$ become different.
My question is:
I'm interested whether there exists a common approach to find the generating function for a given sequence (of numbers). Or perhaps the problem above may be solved in a different way which would give the general term formula without "guessing" its form.
An alternative method is given here (for the equation $x_{m+n}=x_m+x_n+mn$). Let $$f(k):=x_k-\frac{1}{2}\,k^2\text{ for every }k\in\mathbb{Z}_{>0}\,.$$ Then, $f$ satisfies Cauchy's functional equation: $$f(m+n)=f(m)+f(n)\text{ for all }m,n\in\mathbb{Z}_{>0}\,.$$ It is well known that the solution is given by $f(k)=\lambda\,k$, where $\lambda$ is a fixed constant. Therefore, $$x_k=\lambda\,k+\frac{1}{2}\,k^2\text{ for }k=1,2,\ldots\,.$$ As $x_1=a$, we get $\lambda+\dfrac12=a$, whence $$x_k=\left(a-\frac{1}{2}\right)\,k+\frac{1}{2}\,k^2=a\,k+\frac{k(k-1)}{2}\text{ for each }k=1,2,3,\ldots\,.$$
For the equation $x_{m+n}=x_m+x_n+m+n$, we note that $x_2=2x_1+2$ and so $$x_3=x_1+x_2+3=x_1+(2x_1+2)+3=3x_1+5\,.$$ Thus, $$x_4=2x_2+4=4x_1+8$$ but $$x_4=x_1+x_3+4=x_1+(3x_1+5)+4=4x_1+9\,,$$ that is, $4x_1+8=x_4=4x_1+9$, which is a contradiction. Therefore, such a sequence does not exist.
First, we shall prove the converse. Define $$h(k):=\sum_{s=1}^{k-1}\,g(s,1)\text{ for }k=1,2,3,\ldots\,.$$ From (*), it follows immediately that $$h(m+n)=h(m)+h(n)+g(m,n)\text{ for each }m,n\in\mathbb{Z}_{>0}\,.$$ Thus, the function $f:\mathbb{Z}_{>0}\to\mathbb{C}$ defined by $f(k)=x_k-h(k)$ for every $k=1,2,3\ldots$ satisfies Cauchy's functional equation. That is, $f(k)=\lambda\,k$ for some fixed constant $\lambda\in\mathbb{C}$. Hence, $$x_k=\lambda\,k+h(k)=\lambda\,k+\sum_{s=1}^{k-1}\,g(s,1)\text{ for any }k=1,2,3,\ldots\,.$$ The value $x_1$ determines the whole sequence (noting that $\lambda=x_1$).
For the direct implication, we note that $$x_{s+1}=x_s+x_1-g(s,1)\text{ for every }s=1,2,3,\ldots\,.$$ That is, $$x_k-k\,x_1=\sum_{s=1}^{k-1}\,\big(x_{s+1}-x_s-x_1\big)=\sum_{s=1}^{k-1}\,g(s,1)\text{ for all }k=1,2,3,\ldots\,.$$ Thus, for each $k\in\mathbb{Z}_{>0}$, we obtain $$x_k=k\,x_1+h(k)\,,\text{ where }h(k):=\sum_{s=1}^{k-1}\,g(s,1)\,.$$ Since $x_{m+n}=x_m+x_n+g(m,n)$, we finally obtain $$h(m+n)=h(m)+h(n)+g(m,n)\text{ for all }m,n\in\mathbb{Z}_{>0}\,,$$ which is equivalent to (*).
Remark: For a sequence $\left(x_k\right)_{k\in\mathbb{Z}_{>0}}$ to exist, it is necessary that $g$ is symmetric in its two arguments. It is important to note that, if $g$ satisfies (*), then $g$ is a symmetric function.
We can apply the general result to the OP's problem. For the equation $x_{m+n}=x_m+x_n+mn$, we have $g(m,n)=mn$. Therefore, $$\sum_{s=m}^{m+n-1}\,g(s,1)-\sum_{s=1}^{n-1}\,g(s,1)=n\left(m+\frac{n-1}{2}\right)-\frac{n(n-1)}{2}=mn=g(m,n)\,.$$ Therefore, there exists such a sequence, and it is given by $$x_k=k\,x_1+\sum_{s=1}^{k-1}\,g(s,1)=a\,k+\frac{k(k-1)}{2}\,.$$
For the equation $x_{m+n}=x_m+x_n+m+n$, we obtain $g(m,n)=m+n$. Note that $$\sum_{s=m}^{m+n-1}\,g(s,1)-\sum_{s=1}^{n-1}\,g(s,1)=n\left(m+\frac{n+1}{2}\right)-\frac{(n-1)(n+2)}{2}=mn+1\,.$$ Thus, (*) is not satisfied, and such a sequence cannot exist.