<s>[INST] How to convert this $k$th term to Telescopic series Given $$t_k=k^4-2k^3+2k^2-1$$ we need to find
$$S=\sum_{k=1}^{k=10}t_k$$
I factorized $t_k$ as
$$t_k=k(k-1)(k^2-k+1)$$ which further i reduced it as
$$t_k=(k^2-k)^2+(k^2-k)$$
Can i have any hint to reduce this in form $$F(k+1)-F(k)$$
 [/INST] It is more practical to convert $t_k$ into a linear combination of binomial coefficients, then exploit the hockey stick identity. We have
$$ t_k = k^4-2k^3+2k^2-1 = 24\binom{k+1}{4}+6\binom{k}{2}+\binom{k-1}{1} \tag{1}$$
hence:
$$ \sum_{k=1}^{10}t_k = 24\binom{12}{5}+6\binom{11}{3}+\binom{10}{2}=\color{red}{20043}.\tag{2}$$
As a side effect, we also get that $t_k$ can be expressed in terms of the backward differences of $\frac{k^5}{5}+\frac{k^2}{2}-\frac{7k}{10}=24\binom{k+2}{5}+6\binom{k+1}{3}+\binom{k}{2}$.
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