Statement The identity Ln−1+Ln+1=5FnL_{n-1}+L_{n+1}=5F_nLn−1+Ln+1=5Fn holds. Verify it for n=6n=6n=6. Solution With n=6n=6n=6: L5=11L_5=11L5=11 and L7=29L_7=29L7=29, so L5+L7=11+29=40=5⋅8=5F6.L_5+L_7=11+29=\boxed{40}=5\cdot 8=5F_6.L5+L7=11+29=40=5⋅8=5F6.