Finding the value of $n$ in this question?

Question

Find the value of n for which the quadratic equation $$ \sum_{k=1}^{n}(x+k-1)(x+k) =10n $$ has solutions $α$ and $α+1$ for some $α$.

Answer 1

Hint

If you expand the sum appearing in the lhs, you have $$\sum_{k=1}^{n}(x+k-1)(x+k) =\frac{n}{3} \left(n^2+3 n x+3 x^2-1\right)$$ whish is quadratic equation for which you have conditions to be fulfilled.

I am sure that you can take from here.0

Added later to this answer

If you expand the term in the summation, you have $$(x+k-1)(x+k) =x^2-x+2 k x-k+k^2$$ Then $$S=\sum_{k=1}^{n}(x+k-1)(x+k) =\sum_{k=1}^{n}x^2-\sum_{k=1}^{n}x+(2x-1)\sum_{k=1}^{n}k+\sum_{k=1}^{n}k^2$$ that is to say $$S=nx^2-nx+(2x-1)\sum_{k=1}^{n}k+\sum_{k=1}^{n}k^2$$ but (Faulhaber's formulas) $$\sum_{k=1}^{n}k=\frac{1}{2} n (n+1)$$ and $$\sum_{k=1}^{n}k^2=\frac{1}{6} n (n+1) (2 n+1)$$ and then the result.