MATH 236: Discrete Mathematics with Applications
TUTORIAL 8: 2 May, 2013
1. Let P (n) be the statement that
12 + 22 + · · · + n2 = n(n + 1)(2n + 1)/6
for the positive integer n.
(a) What is the statement P (1)?
(b) Show that P (1) is true, completing the basis step of the proof.
(c) What is the inductive hypothesis?
(d) What do you need to prove in the inductive step?
(e) Complete the inductive step, identifying where you use the inductive hypothesis.
Solution.
(a) P (1) is the statement that 12 = 1(1+1)(2·1+1)
6
2·3
6
2
(b) Clearly 1(1+1)(2·1+1)
=
=
=
1
=
1
so P (1) is true
6
6
6
(c) The inductive hypothesis is P (k) is true for some k ≥ 1, that is,
k(k + 1)(2k + 1)
6
(d) We need to prove that P (k + 1) is true, that is,
12 + 22 + · · · + k 2 =
12 + 22 + · · · + k 2 + (k + 1)2 =
(k + 1)((k + 1) + 1)(2(k + 1) + 1)
6
(e) We have
12 + 22 + · · · + k 2 + (k + 1)2
= (12 + 22 + · · · + k 2 ) + (k + 1)2
+ (k + 1)2 by induction hypothesis
= k(k+1)(2k+1)
6
k(2k+1)
= (k + 1)( 6
+ 6(k+1)
)
6
2
2k +7k+6
= (k + 1)
6
= (k+1)(k+2)(2k+3)
.
6
Thus P (k + 1) is true and hence by mathematical induction, P (n) is true for all n ∈ N.
2. For a nonnegative integer n, prove that
3 + 3 · 5 + 3 · 52 + · · · + 3 · 5n = 3(5n+1 − 1)/4.
Solution.
1
• Basis step: We have 3 = 3(5 4−1) so the statement is true when n = 0.
• Assume that the statement is true for n = k ≥ 0. Then
3(5k+1 − 1)
.
4
• We will prove that the statement is true for k + 1. We have
3 + 3 · 5 + 3 · 52 + · · · + 3 · 5k =
3 + 3 · 5 + 3 · 52 + · · · + 3 · 5k + 3 · 5k+1
= (3 + 3 · 5 + 3 · 52 + · · · + 3 · 5k ) + 3 · 5k+1
k+1
= 3(5 4 −1) + 3 · 5k+1
k+1
k+1
= 3( 5 4 −1 + 4·54 )
k+1
= 3 5·5 4 −1
k+2
= 3(5 4 −1)
So the statement is true for n = k + 1 so that by mathematical induction, the statement above
is true for n ∈ N.
1
2
3. Let n be a nonnegative integer. Prove that
2 − 2 · 7 + 2 · 72 − · · · + 2(−7)n = (1 − (−7)n+1 )/4.
Solution.
1
)
• Since 2 = (1−(−7)
, the statement is true for n = 0.
4
• Assume that
2 − 2 · 7 + 2 · 72 − · · · + 2(−7)k =
1 − (−7)k+1
.
4
• Then
2 − 2 · 7 + 2 · 72 − · · · + 2(−7)k + 2(−7)k+1
= (2 − 2 · 7 + 2 · 72 − · · · + 2(−7)k ) + 2(−7)k+1
k+1
= 1−(−7)
+ 2(−7)k+1
4
1−(−7)k+1 +8(−7)k+1
=
4
k+1
= 1+(−7)
4
k+2
= 1−(−7)
4
So the statement is true for n = k + 1 so that by mathematical induction, the statement above
is true for n ∈ N.
4. Answer the following.
(a) Find a formula for
1
1
1
+
+ ··· +
1·2 2·3
n(n + 1)
by examining the values for this expression for small values of n
(b) Prove the formula you conjectured in part (a).
Solution.
(a) Computing the sum on the left side for various values of n, we guess that
1
1
1
1
+
+ ··· +
=1−
.
1·2 2·3
n(n + 1)
n+1
1
1
(b) Using mathematical induction, if n = 1 then 1·2
= 12 = 1 − 1+1
so the statement is true when
n = 1. Assume that
1
1
1
1
+
+ ··· +
=1−
.
1·2 2·3
k(k + 1)
k+1
Then
1
1·2
5. Find
(1)
(2)
(3)
+
1
2·3
+ ··· +
1
(k+1)(k+2)
=
=
=
=
=
1
( 1·2
+
1
2·3
1
1
+ · · · + k(k+1)
) + (k+1)(k+2)
1
1
(1 − k+1 ) + (k+1)(k+2)
k+2
1
+ (k+1)(k+2)
1 − (k+1)(k+2)
k+1
1 − (k+1)(k+2)
1
1 − k+2
.
So the statement is true for n = k + 1 so that by mathematical induction, the statement above
is true for n ∈ N.
f (1), f (2), f (3) and f (4) if f (n) is defied recursively by f (0) = 1 and for n ≥ 1,
f (n + 1) = f (n) + 2
f (n + 1) = 3f (n)
f (n + 1) = f (n)2 + f (n) + 1
3
Solution. Assume that f (n + 1) = f (n) + 2 with f (0) = 1. Then
f (1) = f (0) + 2 = 1 + 2 = 3
f (2) = f (1) + 2 = 3 + 2 = 5
f (3) = f (2) + 2 = 5 + 2 = 7
f (4) = f (3) + 2 = 7 + 2 = 9
Other questions are similar.
6. Find f (2), f (3), f (4) and f (5) if f (n) is defined recursively by f (0) = f (1) = 1 and for n ≥ 1,
(1) f (n + 1) = f (n) − f (n − 1)
(2) f (n + 1) = f (n)f (n − 1)
(3) f (n + 1) = f (n)2 + f (n − 1)3
7. Give a recursive definition of the sequence {an }, n = 1, 2, 3, · · · , if
(1) an = 6n
(2) an = 2n + 1
(3) an = 4n − 2
(4) an = 1 + (−1)n
(5) an = n2 .
Solution.
(1) Since an = 6n, we deduce that an+1 = 6(n + 1) = 6n + 6 = an + 6. Hence, the recurrence relation
is an+1 = an + 6 with a0 = 0.
(2) Since an = 2n + 1, we have an+1 = 2(n + 1) + 1 = 2n + 3 = (2n + 1) + 2 = an + 2. Hence
an+1 = an + 2 with a0 = 1.
(3) As an = 4n − 2, we have an+1 = 4(n + 1) − 2 = 4n + 2 = (4n − 2) + 4 = an + 4 with a0 = −2.
(4) As an = 1 + (−1)n , we have an+1 = 1 + (−1)n+1 = 1 − (−1)n = 1 − (an − 1) = 2 − an since
(−1)n = an − 1. Thus an+1 = 2 − an with a0 = 2
(5) Since an = n2 , we have an−1 = (n − 1)2 = n2 − 2n + 1 and an+1 = (n + 1)2 = n2 + 2n + 1. So
an+1 + an−1 = 2n2 + 2 = 2an + 2.
Thus an+1 = 2an − an−1 + 2 with n ≥ 1 with initial conditions a0 = 0 and a1 = 1.
9. Solving the following recurrence relations together with the initial conditions given.
(1) an = 2an−1 for n ≥ 1 and a0 = 3
(2) an = an−1 for n ≥ 1 and a0 = 2
(3) an = 5an−1 − 6an−2 for n ≥ 2, a0 = 1 and a1 = 0
(4) an = 4an−1 − 4an−2 for n ≥ 2, a0 = 6 and a1 = 8
Solution.
(1) We have that a1 = 2a0 = 2 · 3, a2 = 2a1 = 2 · 2 · 3 = 3 · 22 and so on we can guess that an = 3 · 2n .
Now using induction, we prove that actually an = 3 · 2n . (Prove this)
(2) Clearly, we can see that an = an−1 = · · · = a1 = a0 = 2.
(3) For this case, the characteristic equation is
x2 − 5x + 6 = 0.
Solving this equation, we obtain that r1 = 2 and r2 = 3. Since r1 6= r2 , we obtain that
an = α1 r1n + α2 r2n = α1 2n + α2 3n .
Since a0 = 1 and a1 = 0, we deduce that
a0 = 1 = α1 + α2
a1 = 0 = 2α1 + 3α2
4
Hence,
α1 = 3
α2 = −2.
Therefore,
an = 3 · 2n − 2 · 3n .
(4) The characteristic equation is
x2 − 4x + 4 = 0.
Solving this equation, we obtain that r1 = r2 = 2. Since r1 = r2 , we obtain that
an = α1 rn + α2 nrn = α1 2n + α2 n · 2n .
Since a0 = 6 and a1 = 8, we deduce that
a0 = 6 = α1
a1 = 8 = 2α1 + 2α2
Hence,
α1 = 6
α2 = −2.
Therefore,
an = 6 · 2n − 2n · 2n .
10. Solving the following recurrence relations together with the initial conditions given.
(1) an = an−1 + 6an−2 for n ≥ 2 and a0 = 3, a1 = 6
(2) an = 7an−1 − 10an−2 for n ≥ 2 and a0 = 2, a1 = 1
(3) an = 2an−1 − an−2 for n ≥ 2, a0 = 4 and a1 = 1
(4) an = −6an−1 − 9an−2 for n ≥ 2, a0 = 3 and a1 = −3
11. Consider the nonhomogeneous linear recurrence relation
an = 3an−1 + 2n .
(1) Show that an = −2n+1 is a solution of this recurrence relation.
(2) Use Theorem 3 to find all solutions of this recurrence relation.
(3) Find the solution with a0 = 1.
Solution.
(1) We have 3an−1 + 2n = 3(−2n ) + 2n = −2 · 2n = −2n+1 so an = −2n+1 is a solution of this
(p)
recurrence relation. We denote this solution by an .
(h)
(2) We need to find a solution an of the homogeneous recurrence relation an = 3an−1 . Using
(h)
mathematical induction, we obtain that an = α3n for some real number α.
(p)
(h)
(3) Therefore, an = an + an = α3n − 2n+1 are all solutions of this recurrence relation.
(4) Assume that a0 = 1. Then
a0 = α · 30 − 20+1 = α − 2 = 1
so that α = 3 and hence an = 3 · 3n − 2n+1 = 3n+1 − 2n+1 is the solution of this recurrence
relation with a0 = 1.
12. Consider the nonhomogeneous linear recurrence relation
an = 2an−1 + 2n .
(1) Show that an = n2n is a solution of this recurrence relation.
(2) Use Theorem 3 to find all solutions of this recurrence relation.
(3) Find the solution with a0 = 2.