If $f(t) = 1+ \frac{1}{2} +\frac{1}{3}+....+\frac{1}{t}$, find $\sum^n_{r=1} (2r+1)f(r)$ in terms of $f(n)$ If $f(t) = 1+ \frac{1}{2} +\frac{1}{3}+....+\frac{1}{t}$,
Find $x$ and $y$ such that $\sum^n_{r=1} (2r+1)f(r) =xf(r) -y$

Since $(n+1)^2 - n^2 = 2n+1$, we'd expect $\sum_{r=1}^n (2r+1)f(r)$ to be something like $n^2 f(n)$. A little experimentation shows that it's actually $(n+1)^2 f(n) - n(n+1)/2$. $\textbf{Proof}$: At $n=1$, this is $4 - 1 = 3 = (2+1)f(1)$. 
The forward difference of $(n+1)^2 f(n) - n(n+1)/2$ is $(n+2)^2 f(n+1) - (n+1)(n+2)/2 - [(n+1)^2 f(n) - n(n+1)/2] 
= (n+2)^2 f(n+1) - (n+1)^2 (f(n+1)-1/(n+1)) - (n+1) 
= (2n+3)f(n+1) = (2(n+1)+1)f(n+1)$,
as required.
