Limit of a function in terms of exponents 
Let $f(n)$ denote the $n$th term of the sequence $2, 5, 10, 17, 26,\cdots$ and $g(n)$ denote the $n$th term of the
sequence $2, 6, 12, 20,30,\cdots$.
Let $F(n)$ and $G(n)$ denote respectively the sum of n terms of the above sequences.

$$\lim_{n\to\infty}\left(\frac{F(n)}{G(n)}\right)^n-\left(\frac{f(n)}{g(n)}\right)^n$$
I found

*

*$G(n)=\frac {n(n+1)(n+2)}{3}$

*$F(n)=\frac {n(2n^2+3n+7)}{6}$

*$f(n)=n^2+1$

*$g(n)=n^2+n$
I am not able to further calculate the limit. The answer is $e^{-3/2}-e^{-1}$.

The expressions you got for $F(n),G(n),f(n),g(n)$ are correct.
Now,
$\displaystyle \begin{align} \lim_{n \to\infty}\left(\frac{F(n)}{G(n)}\right)^n-\left(\frac{f(n)}{g(n)}\right)^n &= \lim_{n \to\infty}\left(\frac1{2} \frac{2n^2+3n+7}{n^2+3n+2}\right)^n-\left(\frac{n^2+1}{n^2+n}\right)^n\\
&= \lim_{n \to\infty}\left(\frac{n^2+\frac{3n}{2}+\frac{7}{2}}{n^2+3n+2}\right)^n- \lim_{n \to \infty}\left(\frac{1+\frac1{n^2}}{1+\frac1{n}}\right)^n\\
&=\frac{ \lim\limits_{n \to\infty}\left({1+\frac{3}{2n}+\frac{7}{2n^2}}\right)^n}{\lim\limits_{n \to\infty}\left({1+\frac{3}{n}+\frac{2}{n^2}}\right)^n} - \frac{ \lim\limits_{n \to\infty}\left({1+\frac{1}{n^2}}\right)^n}{\lim\limits_{n \to\infty}\left({1+\frac{1}{n}}\right)^n} \\
& \text{(All the limits are now in $1^\infty$ form)} \\
&=\frac{e^{ \lim\limits_{n \to\infty}\left({\frac{3}{2n}+\frac{7}{2n^2}}\right)n}}{e^{ \lim\limits_{n \to\infty}\left({\frac{3}{n}+\frac{2}{n^2}}\right)n}} - \frac{e^{ \lim\limits_{n \to\infty}\left({\frac1{n^2}}\right)n}}{e^{ \lim\limits_{n \to\infty}\left({\frac{1}{n}}\right)n}} \tag{1} \label{1}\\
&=\frac{e^{ \lim\limits_{n \to\infty}\left({\frac{3}{2}+\frac{7}{2n}}\right)}}{e^{ \lim\limits_{n \to\infty}\left({3+\frac{2}{n}}\right)}} - \frac{e^{ \lim\limits_{n \to\infty}\left({\frac1{n}}\right)}}{e^{ \lim\limits_{n \to\infty} 1}} \\
&=\frac{e^{\frac{3}{2}}}{e^3} - \frac{e^0}{e}\\
&=e^{\frac{-3}{2}} -e^{-1} \end{align}$
where in (\ref{1}) we used $\lim\limits_{n\to \infty} {f_1(n)}^{f_2(n)} = e^{\lim\limits_{n\to \infty}({f_1(n)-1}){f_2(n)}}$ since $f_1(n) \to 1,\; f_2(n) \to \infty$ as $n \to \infty$
