Compute n=1∑∞FnFn+2Fn+1,where Fn denotes the nth Fibonacci number, so F0=0 and F1=1.
Solution — click to reveal
Since Fn+1=Fn+2−Fn,FnFn+2Fn+1=FnFn+2Fn+2−Fn=Fn1−Fn+21.Then n=1∑∞FnFn+21=(F11−F31)+(F21−F41)+(F31−F51)+⋯=F11+F21=2.