MATH Section 5.3

Let P(n) be the statement f21 + f22 + ...+ f2n = fn⋅fn+ 1f12⁢ + f22⁢ + ...⁢+ fn2⁢ = fn⋅fn⁢+ 1 , where fnfn is the nth Fibonacci number. Identify the inductive step. (You must provide an answer before moving to the next part.) Multiple Choice Assume f12 + f22 + ... + fk2 = fk + 1 fk + 1 for any arbitrary integer k ≥ 0. Assume f12 + f22 + ... + fk2 = fk - 1 fk + 1 for some arbitrary integer k ≥ 0. Assume f12 + f22 + ... + fk2 = fk + 1 fk + 1 for any arbitrary integer k > 0. Assume f12 + f22 + ... + fk2 = fk fk + 1 for some arbitrary integer k > 0. Explanation The assumption that P(k) is true for some k is called the inductive hypothesis. In this question, the inductive step is f12 + f22 + ... + fk2 = fk fk + 1 for some arbitrary integer k > 0. Solution Assume f12 + f22 + ... + fk2 = fk fk + 1 for any arbitrary integer k > 0.

Assume f12 + f22 + ... + fk2 = fk fk + 1 for some arbitrary integer k > 0.

Determine whether each of these proposed definitions is a valid recursive definition of a function f from the set of nonnegative integers to the set of integers. If f is well defined, find a formula for f(n) when n is a nonnegative integer and prove that your formula is valid. f(0) = 0, f(n) = 2f(n - 2) for n ≥ 1 Choose the correct statement. (You must provide an answer before moving to the next part.) Multiple Choice The definition of f is not valid because defining f(1) would require f(-1), which is not available. The definition of f is not valid because f(n) must be defined in terms of f(n - 1), not in terms of f(n - 2). The definition of f is valid. Explanation The definition of f is not valid because defining f(1) would require f(-1), which is not available. Solution The definition of f is not valid because defining f(1) would require f(-1), which is not available.

The definition of f is not valid because defining f(1) would require f(-1), which is not available.

Determine whether each of these proposed definitions is a valid recursive definition of a function f from the set of nonnegative integers to the set of integers. If f is well defined, find a formula for f(n) when n is a nonnegative integer and prove that your formula is valid. f(0) = 2, f(1) = 3; f(n) = f(n - 1) - 1 for n ≥ 2 Choose the correct statement. (You must provide an answer before moving to the next part.) Multiple ChoiceThe definition of f is not valid because f(2) has not been defined. The definition of f is valid. The definition of f is not valid because it includes more than one base case. The definition of f is not valid because the formula given for f(n) does not holds the value given for f(1). Explanation The given statement is valid. We can solve f(1) using the given initial conditions. Solution The definition of f is valid.

The definition of f is valid.

Give a recursive definition of the sequence {an}, n = 1, 2, 3,... if an = 4n - 2 Multiple Choice a1 = 2 and an + 1 = an + 4 for all n ≥ 0 a1 = 2 and an + 1 = an + 4 for all n ≥ 1 a1= 2 and an + 1 = an - 4 for all n ≥ 1 a1 = 2 and an + 1 = an + 4 + 1 for all n ≥ 1 Explanation Each term is 4 more than the term before it. Solution a1 = 2 and an + 1 = an + 4 for all n ≥ 1

a1 = 2 and an + 1 = an + 4 for all n ≥ 1

Give a recursive definition of the sequence {an}, n = 1, 2, 3,... if an = n(n + 1) Multiple Choice a1 = 2, an = an - 1 + 2n, n > 0 a1 = 2, an = an - 1 + (2n + 1), n > 1 a1 = 2, an = an - 1 + 2n, n > 2 a1 = 2, an = an - 1 + 2n, n > 1 an = an - 1 + 2n, n > 1 Explanation The sequence starts out 2, 6, 12, 20, 30, and so on. The differences between successive terms are 4, 6, 8, 10, and so on. Thus the nth term is 2n greater than the term preceding it. Solution a1 = 2, an = an - 1 + 2n, n > 1

a1 = 2, an = an - 1 + 2n, n > 1

Let P(n) be the statement f21 + f22 + ...+ f2n = fn⋅fn+ 1f12⁢ + f22⁢ + ...⁢+ fn2⁢ = fn⋅fn⁢+ 1 , where fnfn is the nth Fibonacci number. Click and drag expressions to show in algebraic detail that ∀∀ k(P(k) → P(k + 1)) is true. Explanation ∀k(f21+f32+⋯+f2k+f2k+1=(f21+f32+⋯+f2k)+f2k+1∀k(f12+f23+⋯+fk2+fk+12=(f12+f23+⋯+fk2)+fk+12 fkfk+1+f2k+1=fk+1(fk+fk+1)=fk+1f(k+1)+1)

correct blocks and order (going down): (f21+f32+⋯+f2k)+f2k+1 fkfk+1+f2k+1 fk+1(fk+fk+1) fk+1f(k+1)+1)

Determine whether each of these proposed definitions is a valid recursive definition of a function f from the set of nonnegative integers to the set of integers. If f is well defined, find a formula for f(n) when n is a nonnegative integer and prove that your formula is valid. Choose the correct formula for f(n) when n is a nonnegative integer. (You must provide an answer before moving to the next part.) Multiple Choice f(0) = 2; f(n) = n - 4 for n > 0 f(0) = 2; f(n) = 4 - n for n > 0 f(n) = 2n + 2 for n ≥ 0 f(n) = n + 2 for n ≥ 0 Explanation The correct formula for f(n) when n is a nonnegative integer is f(n) = 4 - n for n > 0 where f(0) = 2. Solution f(0) = 2; f(n) = 4 - n for n > 0

f(0) = 2; f(n) = 4 - n for n > 0

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n + 1) = f(n) + 3f(n − 1) for n = 1, 2, .. f(2) = (You must provide an answer before moving to the next part.) Explanation f(2)=f(1)+3f(0)=2+3(−1)=−1f(2)=f(1)+3f(0)=2+3(-1)=-1 Solution f(2) = -1

f(2) = -1

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n+1)=f(n)2f(n− 1)f(n+1)=f(n)2f(n⁢− 1) for n = 1, 2, ... f(2) = (You must provide an answer before moving to the next part.) Explanation f(2)=f(1)2f(0)=(2)2(−1)=−4f(2)=f(1)2f(0)=(2)2(-1)=-4 Solution f(2) = -4

f(2) = -4

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n+1)=f(n)2f(n− 1)f(n+1)=f(n)2f(n⁢− 1) for n = 1, 2, ... f(3) = (You must provide an answer before moving to the next part.) Explanation f(3)=f(2)2f(1)=(−4)2(2)=32f(3)=f(2)2f(1)=(-4)2(2)=32 Solution f(3) = 32

f(3) = 32

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n + 1) = f(n) + 3f(n − 1) for n = 1, 2, .. f(3) = (You must provide an answer before moving to the next part.) Explanation f(3)=f(2)+3f(1)=−1+3(2)=5f(3)=f(2)+3f(1)=-1+3(2)=5 Solution f(3) = 5

f(3) = 5

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n+1)=f(n)2f(n− 1)f(n+1)=f(n)2f(n⁢− 1) for n = 1, 2, ... f(4) = (You must provide an answer before moving to the next part.) Explanation f(4)=f(3)2f(2)=(32)2(−4)=−4096f(4)=f(3)2f(2)=(32)2(-4)=-4096 Solution f(4) = -4096

f(4) = -4096

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n + 1) = f(n) + 3f(n − 1) for n = 1, 2, .. f(4) = (You must provide an answer before moving to the next part.) Explanation f(4)=f(3)+3f(2)=5+3(−1)=2f(4)=f(3)+3f(2)=5+3(-1)=2 Solution f(4) = 2

f(4) = 2

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n + 1) = f(n) + 3f(n − 1) for n = 1, 2, .. f(5) = (You must provide an answer before moving to the next part.) Explanation f(5) = f(4) + 3(f(3))=f(3)+3f(2)=5+3(−1)=2f(4)=f(3)+3f(2)=5+3(-1)=2 Solution f(5) = 17

f(5) = 17

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = −1, f(1) = 2, and f(n+1)=f(n)2f(n− 1)f(n+1)=f(n)2f(n⁢− 1) for n = 1, 2, ... f(5) = Explanation f(5)=f(4)2f(3)=(−4096)2(32)=536,870,912f(5)=f(4)2f(3)=(-4096)2(32)=536,870,912 Solution f(5) = 536,870,912

f(5) = 536,870,912

Let P(n) be the statement f21 + f22 + ...+ f2n = fn⋅fn+ 1f12⁢ + f22⁢ + ...⁢+ fn2⁢ = fn⋅fn⁢+ 1 , where fnfn is the nth Fibonacci number. Identify the basis step to prove that P(n) is true. (You must provide an answer before moving to the next part.) Multiple Choice f20 = f0⋅f1f02⁢ = f0⋅f1 = 1 at n = 1 f21 = f1⋅f2f12⁢ = f1⋅f2 = 1 at n = 1 f21 = f1⋅f2f12⁢ = f1⋅f2 = 0 at n = 0 f21 = f1⋅f2f12⁢ = f1⋅f2 = 1 at n = 2 f21 = f1⋅f2f12⁢ = f1⋅f2 = 1 at n = 0 Explanation We will have completed the basis step of the proof if we show that P(1) is true because f21 = f1⋅f2f12⁢ = f1⋅f2 = 1 is true for n = 1. Solution f21 = f1⋅f2f12⁢ = f1⋅f2 = 1 at n = 1 is the basis step.

f21 = f1⋅f2f12⁢ = f1⋅f2 = 1 at n = 1