Book of Proof - Amazon S3

280 Solutions
23. Use induction to prove the binomial theorem (x + y) n = ∑ (
n n
i=0 i)
x n−i y i .
Proof. Notice that when n = 1, the formula is (x + y) 1 = ( 1) 0 x 1 y 0 + ( 1
1)
x 0 y 1 = x + y,
which is true.
Now assume the theorem is true for some n > 1. We will show that this implies
that it is true for the power n + 1. Just observe that
(x + y) n+1 = (x + y)(x + y) n
)
n∑ n
= (x + y) x
i=0( n−i y i
i
)
)
n∑ n n∑ n
= x
i=0( (n+1)−i y i + x
i
i=0( n−i y i+1
i
) ( )]
n∑ n n
= + x
i=0[( (n+1)−i y i + y n+1
i i − 1
( )
( )
n∑ n + 1
n + 1
=
x (n+1)−i y i + y n+1
i=0
i
n + 1
( )
n+1 ∑ n + 1
=
x (n+1)−i y i .
i
i=0
This shows that the formula is true for (x + y) n+1 , so the theorem is proved.
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25. Concerning the Fibonacci sequence, prove that F 1 +F 2 +F 3 +F 4 +...+F n = F n+2 −1.
Proof. The proof is by induction.
(1) When n = 1 the statement is F 1 = F 1+2 − 1 = F 3 − 1 = 2 − 1 = 1, which is true.
Also when n = 2 the statement is F 1 + F 2 = F 2+2 −1 = F 4 −1 = 3−1 = 2, which is
true, as F 1 + F 2 = 1 + 1 = 2.
(2) Now assume k ≥ 1 and F 1 + F 2 + F 3 + F 4 + ... + F k = F k+2 − 1. We need to show
F 1 + F 2 + F 3 + F 4 + ... + F k + F k+1 = F k+3 − 1. Observe that
F 1 + F 2 + F 3 + F 4 + ... + F k + F k+1 =
(F 1 + F 2 + F 3 + F 4 + ... + F k ) + F k+1 =
F k+2 − 1 + +F k+1 = (F k+1 + F k+2 ) − 1
= F k+3 − 1.
This completes the proof by induction.
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27. Concerning the Fibonacci sequence, prove that F 1 + F 3 + ··· + F 2n−1 = F 2n .
Proof. If n = 1, the result is immediate. Assume for some n > 1 we have
∑ n
i=1 F 2i−1 = F 2n . Then ∑ n+1
i=1 F 2i−1 = F 2n+1 + ∑ n
i=1 F 2i−1 = F 2n+1 +F 2n = F 2n+2 = F 2(n+1)
as desired.
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