# Proving using induction or strong induction on Fibonacci number proposition [duplicate]

I'm stuck on proving this statement below due to that I can't seem to find a base case that is true:

Prove that $$\sum_{i=0}^{2n}(-1)^if(i) = f(0) - f(1) + f(2) - \cdots - f(2n-1) + f(2n) = f(2n-1)-1,$$ where $f(i)$ is the $i$th Fibonacci number.

I believe simple induction should work but I am unsure if I have to use Strong Induction.

## marked as duplicate by lulu, Namaste discrete-mathematics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 21 '17 at 0:08

• For what $n$ do you need to prove this? – Mastrem Mar 20 '17 at 20:09
• The sum looks like it should be $\sum (-1)^iF(i)$ not $F(2i)$. – Mark Fischler Mar 20 '17 at 20:11
• why you are "unsure" about using weak or strong induction? Just try the weak induction approach, if it gets complicate try the other. So you dont need to be unsure! – Masacroso Mar 20 '17 at 20:12

For $n+1$ we have

$$\sum_{i=0}^{2(n+1)}(-1)^if(i)=\sum_{i=0}^{2n}(-1)^if(i)-f(2n+1)+f(2n+2)=\\ =f(2n-1)-1-f(2n+1)+f(2n+2)$$

but

$$f(2n+2)=f(2n+1)+f(2n),\\ f(2n+1)=f(2n)+f(2n-1)$$ so,

$$f(2n-1)-f(2n+1)+f(2n+2)=f(2n+1)$$

and then

$$\sum_{i=0}^{2(n+1)}(-1)^if(i)=f(2n+1)-1=f[2(n+1)-1]-1$$