2020 ACJC P1 Q5

Let \(\mathrm{f}\left( n \right)=\frac{{{3}^{n}}}{n}\), where \(n\) is a positive integer. Show that

\(\mathrm{f}\left( n+1 \right)-\mathrm{f}\left( n \right)=\frac{{{3}^{n}}\left( 2n-1 \right)}{n\left( n+1 \right)}\).[1]

Hence find \(\sum\limits_{n=1}^{N}{\left( \frac{{{3}^{n}}}{n+1}\left( 1-\frac{1}{2n} \right) \right)}\) in terms of \(N\).[4]

Deduce \(\sum\limits_{n=1}^{N}{\frac{{{3}^{n}}\left( 2n+1 \right)}{\left( n+1 \right)\left( n+2 \right)}}\), leaving your answer in the form \(a\left( \frac{{{3}^{N+2}}}{N+2}-b \right)\), where \(a\) and \(b\) are constants to be determined.[3]