(a) Use standard results from MF19 to find \(\sum_{r=1}^{n}(8r^3+12r^2+4r+3)\) in terms of \(n\), simplifying your answer.

(b) Show that \(\frac1{r^4}-\frac1{(r+1)^4}=\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}\), and hence find \(\sum_{r=1}^{n}\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}\).

(c) Deduce the value of \(\sum_{r=1}^{\infty}\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}\).