00:01
All right, so for problem 16, again, we need to apply induction to prove the relation here.
00:11
So the basis step we consider is when n is equal to 1.
00:18
So we have f0 minus f1 plus f2, which is equal to 0 minus 1 plus 1, which is 0.
00:33
F1 minus 1, which is 0.
00:39
Now we assume that when it is equal to k, now we still have f0 minus f1 plus f2 minus dot dot dot minus f 2k minus 1 plus f2k, which is equal to f2k, 2k minus 1 minus 1.
01:12
Now we want to show that when n is equal to k plus 1, this division still holds.
01:20
Now we first write down right down the case when n is equal to k plus 1 on the left side.
01:29
So we have f0 minus f1 plus f2 minus dada -da -dun minus f2k plus 1 plus f 2k plus 2k plus 2k plus 2 and so yeah just for just for clarification i i will write i will write more terms before before f2k plus 1 so that's f0 minus f1 plus f2 minus f2k minus 1 because now we can compare this expression with our previous case when n is equal to k.
02:28
So minus f2k minus 1 plus f 2k minus f 2k plus 1 plus f 2k plus 2...