Find the sum up to $n$ terms where $k_{\text{th}}$ term is $k(k+1)(k²+k+1).$
My attempt: I multiplied and divided by three then wrote $$1/3\cdot((k²+k+2)-(k²+k-1))(k²+k)(k²+k+1)$$ but couldn't succeed to make a telescoping series. Anyone doing by making a telescope would be helpful. Other methods are also welcome.
On
Hint
Let $$g(k)=(k^2+k)(k^2+k+1)=f(k+1)-f(k)$$ where $f(m)=\sum_{r=0}^na_rm^r$
$$\implies\sum_{k=1}^ng(k)=f(n+1)-f(1)$$
Clearly $a_r=0\forall r\ge6$ as the coefficients of $k^r$ is zero for $r\ge6$
Comparing the coefficients of $k^4,k^3,k^2,k,k^0$ one by one
we get
$1=a_5\binom51\iff a_5=?$
$2=\binom52a_5+\binom41a_4\implies a_4=?$
Can you take it from here?
It's always possible to make it (polynomial) into a telescope. E.g. do it term wise for $ \sum k , \sum k^2, \sum k^3 , \sum k^4 \ldots$.
The hard part is figuring out what the $u(k)$ is (without a lot of tedious guessing).
Hint: $(k+1)^5 - k^5 = 5(k)(k+1)(k^2+k+1) +1$ (by expanding it out)
So this should give you a very good guess for $u(k)$.