2022 EJC Promo Q11

It is given that \(\sum\limits_{r=\,\,1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}\).

1. Show that \(\sum\limits_{r=\,\,1}^{n}{\left( 2r-7 \right)\left( r+1 \right)}=\frac{1}{6}n\left( 4{{n}^{2}}-9n-55 \right)\).[4]
2. Find \(\sum\limits_{r=\,\,1}^{n}{{{3}^{-\,r}}}\) in terms of \(n\), and find the least value of \(n\) such that \(\sum\limits_{r=\,\,1}^{n}{\left( 2r-7 \right)}\left( r+1 \right)>\sum\limits_{r=\,\,1}^{n}{{{3}^{-\,r}}}\).[2]
3. Express \(\sum\limits_{r=\,\,n+1}^{2n}{\left( 2r-7 \right)\left( r+1 \right)}\) in terms of \(n\).[2]
4. Hence, or otherwise, find the value of \(43\times 26+45\times 27+47\times 28+...87\times 48+89\times 49\).[2]