1. There are a lot of ``simple'' identities like $$ 1+\dotsb+n = \frac{n(n+1)}{2} $$ or $$ 1^2+\dotsb+n^2 = \frac{n(n+1)(2n+1)}{6} $$ 2. Some include Fibonacci numbers, like in Prove Fibonacci Identity using generating functions or Help to prove $F_n^5+F_{n+1}^5=F_{n+2}[(F_nF_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_nF_{n+1}]$ .
3. There are also some combinatorical identities, e. g. Show that ${n\choose r}2^r 3^{n-r}=\sum_{k=r}^{n} {n \choose k} {k \choose r}2^k$
Identities of the first type can be generated by solving coefficients, e. g. $$ 1^7 + 4^7 + \dotsb + (3n+1)^7 $$ should probably be a polynomial of 8th degree.
I am interested if there is any way of generating identities of the second or the third type.
Third type:
For most usual sums whose terms contain binomial coefficients (including your example), the combination of Zeilberger and Petkovsek algorithms will check whether there is a closed form.
See http://mathworld.wolfram.com/ZeilbergersAlgorithm.html and https://en.wikipedia.org/wiki/Petkov%C5%A1ek%27s_algorithm for more information and references.