Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation fk = fx-1+ 2k for each integer k 2 2 1 = 1 satisfies the following formula. f, = 2" +1- 3 for every integer n 2 1 Proof (by mathematical induction): Suppose f,, f2, f3 is a sequence that satisfies the recurrence relation 1' ... Ik = k + 2k for each integer k 2 2, with initial condition f, = 1. We need to show that when the sequence f,, f,, f3, ... is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above. So let the property P(n) be the equation f, = 2" +1 - 3. We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: The left-hand side of P(1) is which equals The right-hand side of P(1) is Since the left-hand
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![Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation
fk = fk
+ 2k for each integer k 2 2
- 1
= 1
satisfies the following formula.
f, = 2" +1 - 3 for every integer n 2 1
Proof (by mathematical induction):
Suppose f,, f2, f31
... is a sequence that satisfies the recurrence relation
f = fk -1+ 2* for each integer k 2 2, with initial condition f, = 1.
We need to show that when the sequence f,, f,, f3, ... is defined in this recursive way, all the terms in the sequence also satisfy the
explicit formula shown above.
So let the property P(n) be the equation f, = 2" +1 - 3. We will show that P(n) is true for every integer n 2 1.
Show that P(1) is true:
The left-hand side of P(1) is
which equals
. The right-hand side of P(1) is
. Since the left-hand
and right-hand sides equal each other, P(1) is true.
Show that for each integer k 2 1, if P(k) is true, then P(k + 1) is true:
Let k be any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f =
, where f,, f2, f3, ... is
a sequence defined by the recurrence relation
f = fx-1+ 2k for each integer k > 2, with initial condition f, = 1.
[This is the inductive hypothesis.]
We must show that P(k + 1) is true. In other words, we must show that
Now the left-hand side of P(k + 1) is
+ 1
fk + 1
k + 2k + 1
by definition of f, f, f, .'
2k + 1
by inductive hypothesis
+
= 2 .
by the laws of algebra
and this is the right-hand side of P(k + 1). Hence the inductive step is complete.
[Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b3df428-2a91-49dd-84ab-611ffd39f8ca%2Fba5955e9-c070-434c-ab16-75f534fb7af7%2Fm8se1lj_processed.png&w=3840&q=75)
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