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The question I had was calculating $$displaystylefrac{1}{k+1-j}sum_{i=j}^k i^2$$

Because I didn't know how to do a variable change, I did
$$frac{1}{k+1-j}sum_{i=j}^k i^2 = frac{1}{k+1-j}left(sum_{i=1}^k i^2 - sum_{i=1}^{j-1} i^2right) = frac{2k^3+3k^2+k-2j^3+3j^2-j}{6(k+1-j)}$$

The solution given used variable change and was able to cancel out $(k+1-j)$ very easily. Finally the answer was $$frac{2k^2 + 2j^2 + 2kj - j + k}{6}$$ I have verified that these two solutions are the same by multiplying the second one with $(k+1-j).$

My question is:

How do you divide an expression like ${(2k^3+3k^2+k-2j^3+3j^2-j)}$ with $(k+1-j)$? Is it even worth it in this case to get the most simplified solution?

You could use synthetic division fairly easily here.

First, we consider $2k^3+3k^2+k-2j^3+3j^2-j$ as a polynomial in $k,$ and write the $4$ coefficients $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\ & & & &\hline & & & &end{array}$$ Next, we invert the coefficients of the divisor $k+1-j$ to get $-1k+j-1,$ and write the constant term in as $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\j-1 & & & &\hline & & & &end{array}$$ We bring down the first term to get $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\j-1 & & & &\hline & 2 & & &end{array}$$ Now, we multiply what we've just brought down by $j-1$ and write in the result as $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\j-1 & & 2j-2 & &\hline & 2 & & &end{array}$$ Now, adding the numbers in that column gets us $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\j-1 & & 2j-2 & &\hline & 2 & 2j+1 & &end{array}$$ Multiply that result by $j-1$ and write it in to get $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\j-1 & & 2j-2 & 2j^2-j-1 &\hline & 2 & 2j+1 & &end{array}$$ Adding again gives us $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\j-1 & & 2j-2 & 2j^2-j-1 &\hline & 2 & 2j+1 & 2j^2-j &end{array}$$ Multiplying again by $j-1$ and then adding once more, we get $$begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\j-1 & & 2j-2 & 2j^2-j-1 & 2j^3-3j^2+j\hline & 2 & 2j+1 & 2j^2-j &0end{array}$$ Translating back into terms of a polynomial in $k,$ this means that $$frac{2k^3+3k^2+k-2j^3+3j^2-j}{k+1-j}=2k^2+(2j+1)k+2j^2-j+frac{0}{k+1-j},$$ or more simply, $$frac{2k^3+3k^2+k-2j^3+3j^2-j}{k+1-j}=2k^2+2j^2+2jk-j+k,$$ as desired.

Try to write the polynomial in $k$ as a polynomial in $m=k+1$.

$2k^3+3k^2+k=big[2(k+1)^3-2-6k-6k^2big]+big[3(k+1)^2-3-6kbig]+(k+1)-1\=2(k+1)^3+3(k+1)^2+(k+1)-6(1+2k+k^2)\=2m^3-3m^2+m$

Therefore, $2m^3-3m^2+m-2j^3+3j^2-j=2(m^3-j^3)-3(m^2-j^2)+m-j\=(m-j)big[2(m^2+j^2+mj)-3(m+j)+1big]$

Divide by $k+1-j=m-j$ and back-substitute $m$ to get,

$$2(m^2+j^2+mj)-3(m+j)+1=2k^2+2j^2+2kj+k-j$$

making $k-j+1=mto j=k-m+1$

and after substitution into

$$
2 k^3 + 3 k^2 + k - 2 j^3 + 3 j^2 - j =m + 6 k m + 6 k^2 m - 3 m^2 - 6 k m^2 + 2 m^3
$$