Find the sum

Yes.. Do you have a solution? Cause I posted it on AOPS but get one.

A user there PMed me saying that he solved it using Maple, which gave him answer as 4/3, but still he didn't had a solution.

\hspace{-16}$ Let $\mathbf{k=m+n}$. Then $\mathbf{n=k-m}$\\\\ and $\mathbf{f(m,n)=3m+n+(m+n)^2=(m+n)+(m+n)^2+2m}$\\\\ $\mathbf{f(m,k-m)=k+k^2+2m}$\\\\ $\mathbf{\sum_{n=0}^{\infty}\;\sum_{m=0}^{\infty}2^{-f(m,n)}=\sum_{k=0}^{\infty}\;\sum_{m=0}^{k}2^{-f(m,k-m)}=\sum_{k=0}^{\infty}\;\sum_{m=0}^{k}2^{-(k+k^2+2m)}}$\\\\\\ $\mathbf{=\sum_{k=0}^{\infty}\;\sum_{m=0}^{k}2^{-2\left(\frac{k(k+1)}{2}+m\right)}=\sum_{k=0}^{\infty}\;\sum_{m=0}^{k}4^{-\left(\frac{k(k+1)}{2}+m\right)}}$\\\\\\ Put $\mathbf{\frac{k(k+1)}{2}+m = p}$\\\\\\

\hspace{-16}\mathbf{=\sum_{k=0}^{\infty}\;\sum_{p=\frac{k(k+1)}{2}}^{\frac{k(k+1)}{2}+k}4^{-p}=\sum_{k=0}^{\infty}\;\sum_{p=\frac{k(k+1)}{2}}^{\frac{(k+1)(k+2)}{2}-1}4^{-p}}$\\\\\\ Now Simplifycing These Two Summation, We Get\\\\ $\mathbf{\sum_{p=0}^{\infty}4^{-p}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}}$

Not Mine Solution.............

also seen in 1 or 2 year back FIITJEE AITS paper..........

THanks..

The solution however doesn't seems too trivial.