MAT 283
EXAM #2 Solutions
2013-03-20
1. Express each of the following sums using summation notation.
(a) (5 points) 4 + 5 + 6 + 7
Solution: One possibility is
7
X
i.
i=4
(b) (7 points)
1
1 1 1
+ + + ··· +
2 4 6
20
10
X
1
Solution: One possibility is
.
2i
i=1
(c) (7 points) 5 + 9 + 13 + 17 + 21 + · · · + 49
Solution: One possibility is
11
X
(4i + 5). Another is
i=0
12
X
(4i + 1).
i=1
2. (8 points) Let a be a nonzero real number. Find the flaw in the following “proof” that
an = 1 for every nonnegative integer n.
“Proof ”. [By Induction] Basis step: a0 = 1.
1
an
Inductive step: an+1 = an · n−1 = 1 · = 1.
a
1
Solution: We need a1 as a basis step since the induction depends on the two values
before n + 1. Since a1 = a, the proof fails.
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MAT 283
Exam #2 Solutions
3. (12 points) Prove that for t ∈ N,
2
2
t(t + 1)
(t + 1) [(t + 1) + 1]
3
+ (t + 1) =
.
2
2
(Hint: Do not use induction.)
Solution: Notice that
2
t(t + 1)
+ (t + 1)3 =
2
!
(t + 1) 2 2
t + 4(t + 1)
2
!
(t + 1) 2 2
=
t + 4t + 4
2
!
2
(t + 1)
[t + 2]2
=
2
!
2
(t + 1)
[(t + 1) + 1]2
=
2
2
(t + 1) [(t + 1) + 1]
=
.
2
Pn
2
4. (12 points)
Prove
that
i=1 (2i − 1) = n for all n ∈ N. For this problem, you cannot
Pn
use that i=1 i = n(n + 1)/2 (since this is problem 6).
Solution: There are 3 different methods for this:
P
Method #1: [By Induction] Base Case: Notice that 1i=1 (2i − 1) = 1 = 12 .
P
Pn
2
Inductive Step: Assume n−1
i=1 (2i − 1) = (n − 1) . Consider
i=1 (2i − 1).
n
X
i=1
(2i − 1) =
n−1
X
(2i − 1) + (2n − 1)
i=1
= (n − 1)2 + (2n − 1)
= n2 − 2n + 1 + (2n − 1)
= n2 .
n
X
Thus, the inductive step holds and so
(2i − 1) = n2 for all n ∈ N.
i=1
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MAT 283
Exam #2 Solutions
Method #2: Let S =
other
1
(2n − 1)
Pn
i=1 (2i
− 1). We list two copies of the sum, one above the
+
3
+
+ (2(n − 1) − 1) +
5
+ ···
(2(n − 2) − 1) + · · ·
+
+
(2n − 1)
1
This shows that
2S = |2n + ·{z
· · + 2n}
n terms
2S = n(2n)
2S = 2n2
S = n2 .
Method #3: Consider 2n2 points in a grid with 2n rows and n columns. Counted
by columns, there are 2n2 points. We can also group the points into two disjoint
subsets with columns of sizes 1, 3, 5, . . ., 2n − 1.
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MAT 283
Exam #2 Solutions
5. (12 points) Prove that 4n > n3 for every positive integer n.
Solution: Base case: When n = 1, 41 > 13 . Also, to make step (?) in the proof
easier to see, 42 = 16 > 23 = 8 and 43 = 64 > 27 = 33 . (Otherwise, it would need
more support.)
Induction step: Method 1: Assume 4n > n3 . Consider (n + 1)3 where n + 1 ≥ 4.
Notice
(n + 1)3 = n3 + 3n2 + 3n + 1
3
3
3
< n3 + n
| +n
{z + n}
(?)
since n ≥ 3
3
= 4n
< 4 · 4n = 4n+1 .
Therefore 4n > n3 for every positive integer n.
Method 2: Assume 4n > n3 . Consider 4n+1 where n + 1 ≥ 4. This means n3 ≥ 3n2 ,
n3 > 3n, and n3 > 1. Notice
4n+1 = 4 · 4n
> 4 · n3 = n3 + n3 + n3 + n3
> n3 + 3n2 + 3n + 1
= (n + 1)3 .
Therefore 4n > n3 for every positive integer n.
6. (12 points) Prove that
n
X
i=
i=1
n(n + 1)
for every positive integer n.
2
Solution: There are 3 different methods for this:
P
Method #1: [By Induction] Base Case: For n = 1, we have 1i=1 i = 1 = n.
P
Pn
2
Inductive Step: Assume n−1
i=1 (2i − 1) = (n − 1) . Consider
i=1 (2i − 1).
n
X
i=1
i=
n−1
X
i+n
i=1
(n − 1)n
=
+n=n
2
Thus, the inductive step holds and so
n
X
n−1
+1
2
i=
i=1
Page 4
=
n(n + 1)
.
2
n(n + 1)
for all n ∈ N.
2
(?)
MAT 283
Exam #2 Solutions
Method #2: Let S =
Pn
1 +
n +
i=1
i. We list two copies of the sum, one above the other
3
(n − 1)
+
5
+ (n − 2)
+ ···
+ ···
+ n
+ 1
This shows that
2S = (n + 1) + · · · + (n + 1)
|
{z
}
n terms
2S = n(n + 1)
n(n + 1)
.
S=
2
Method #3: Consider n(n + 1) points in a grid with n rows and n + 1 columns.
Counted by columns, there are n(n + 1) points. We can also group the points into
two disjoint subsets with columns of sizes 1, 2, 3, . . ., n similar to the image for
n = 9.
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