MAS110 Solutions: Summation and induction
1. Since 12 + 22 + . . . + n2 = n(n + 1)(2n + 1)/6, we have
12 + 22 + . . . + 242 =
24 × 25 × 49
= 702 .
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2. The sum of the first n natural numbers is n(n + 1)/2. So let P (n) be the
statement
n2 (n + 1)2
13 + . . . + n3 =
.
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Base step. P (1) is “13 = 12 .22 /4”, which is true.
Induction step. Assume P (k) is true i.e. 13 + . . . + k 3 = k 2 (k + 1)2 /4. Then
k 2 (k + 1)2
+ (k + 1)3
(by induction hypothesis)
4 k2
= (k + 1)2
+k+1
4
2
k + 4k + 4
= (k + 1)2
4
2
2
(k + 1) (k + 2)
.
=
4
13 + . . . + k 3 + (k + 1)3 =
So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
3. 1 + 3 + . . . + (2n − 1) = n2 .
First proof. If S = 1 + 3 + . . . + (2n − 1) then
2S =
1
+
3
+ . . . + 2n − 3 + 2n − 1 +
2n − 1 + 2n − 3 + . . . +
3
+
1.
So 2S = 2n × n i.e. S = n2 .
Second proof. Use induction. Let P (n) be the statement “1+3+. . .+(2n−1) =
n2 ”.
Base step. 1 = 12 ; so P (1) is true.
Induction step. Assume P (k) is true. Then
1 + . . . + (2k − 1) + (2k + 1) = k 2 + 2k + 1
= (k + 1)
1
2
(by induction hypothesis)
Figure 1: Sum of the first n odd numbers
So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
Third proof. See Figure 1.
4.
(i) Let P (n) be the statement “3 divides n3 − n”.
Base step. 3 does divide 13 − 1; so P (1) is true.
Induction step. Assume P (k) is true i.e. 3 divides k 3 − k. Then
(k + 1)3 − (k + 1) = k 3 + 3k 2 + 3k + 1 − k − 1 = (k 3 − k) + 3k 2 + 3k
shows that 3 divides (k + 1)3 − (k + 1). So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
(ii) Let P (n) be the statement “5 divides n5 − n”.
Base step. 5 does divide 15 − 1; so P (1) is true.
Induction step. Assume P (k) is true i.e. 5 divides k 5 − k. Then
(k+1)5 −(k+1) = k 5 +5k 4 +10k 3 +10k 2 +5k+1−k−1 = (k 5 −k)+5k 4 +10k 3 +10k 2 +5k
shows that 5 divides (k + 1)3 − (k + 1). So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
(iii) Not true: 4 doesn’t divide 24 − 2.
(iv) Let P (n) be the statement “p divides np − n”.
Base step. p does divide 1p − 1; so P (1) is true.
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Induction step Assume P (k) is true i.e. p divides k p − k. Then
p p−1
p
(k + 1) − (k + 1) = k +
k
+ ... +
k+1−k−1
1
p−1
p p−1
p
= (k p − k) +
k
+ ... +
k.
1
p−1
If we can show that the binomial coefficient pj is a multiple of p for
j = 1, 2, . . . , p − 1 then the above expression shows that p divides (k +
1)p − (k + 1) i.e. P (k + 1) is true. We can then conclude that P (n) is
true for all n ∈ N by induction.
For j = 1, 2, . . . , p − 1 , we have
p
p!
=
.
j
j!(p − j)!
p
p
Since the prime p is not a factor of j! or (p − j)!, and p divides p!, it
follows that p must divide the integer pj . This completes the proof.
5. The arithmetic sequence is 11, 11 + d, . . . , 11 + 9d which sums to 11 · 10 +
9(1 + 2 + . . . + 9) i.e. 110 + 45d.
The geometric sequence is 5, 5 · 2, . . . , 5 · 29 . This sums to 5(1 + 2 + . . . + 29 ),
which is 5(210 − 1).
So 110 + 45d = 5(210 − 1), which gives
d=
5 · 210 − 115
1001
=
.
45
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6. Let P (n) be the statement “a + ar + . . . + arn−1 = a(rn − 1)/(r − 1)”.
(r 6= 1, n ∈ N)
Base step. a = a(r − 1)/(r − 1); so P (1) is true.
Induction step. Assume P (k) is true. Then
a(rk − 1)
+ ark
r−1
a((rk − 1) + rk (r − 1))
=
r−1
k+1
a(r
− 1)
=
.
r−1
a + ar + . . . + ark−1 + ark =
So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
3
(by induction hypothesis)
7. Let r be the common ratio. Then b = ar, c = ar2 and d = ar3 . Then
(b − c)2 = (ar − ar2 )2 = a2 r2 (1 − r)2 = a2 r2 − 2a2 r3 + a2 r4 ,
2 2
2 4
and
2 3
ac + bd − 2ad = a r + a r − 2a r .
8. The limit is
110, 000 − 10, 000 + 5, 000 − 2, 500 + . . .
1 1 1
= 110, 000 − 10, 000 1 − + − + . . .
2 4 8
1
= 110, 000 − 10, 000 ×
1 + 12
≈ 103, 333.33.
n
d
λx ) = eλx (λn x + nλn−1 )”.
9. For n ∈ N, let P (n) be the statement “ dx
n (xe
Base step.
d
λx
dx (xe )
= eλx + xλeλx ; so P (1) is true.
Induction step. Assume P (k) is true. Then
k
d k+1
d
d
λx
λx
(xe ) =
(xe )
dx
dx dxk
d λx k
=
e (λ x + kλk−1 )
(by induction hypothesis)
dx
= λeλx (λk x + kλk−1 ) + eλx (λk )
= eλx (λk+1 x + kλk + λk )
= eλx (λk+1 x + (k + 1)kλk ).
So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
10.
(i) For each n ∈ N, let P (n) be the statement “2n+1 < 1 + (n + 1)2n ”.
Base step. 21+1 = 4 and 1 + (1 + 1)21 = 5. So P (1) is true.
Induction step. Assume P (k) is true for some k ∈ N i.e. 2k+1 < 1 + (k +
1)2k . Thus
2(k+1)+1 = 2 · 2k+1 < 2(1 + (k + 1)2k ) = 2 + (k + 1)2k+1 .
Now 2 + (k + 1)2k+1 < 1 + (k + 2)2k+1 ⇐⇒ 1 < 2k+1 , which is true.
So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
(ii) Let x > −1 be fixed, and for each n ∈ N, let P (n) be the statement
“(1 + x)n ≥ 1 + nx”.
Base step. (1 + x)1 = 1 + x ≥ 1 + 1x; so P (1) is true.
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Induction step. Assume P (k) is true for some k ∈ N i.e. (1+x)k ≥ 1+kx.
Now (1 + x)k+1 = (1 + x)(1 + x)k . Hence by the inductive hypothesis,
and since 1 + x > 0, we obtain (1 + x)k+1 ≥ (1 + x)(1 + kx) = 1 + (k +
1)x + kx2 ≥ 1 + (k + 1)x. So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
11. For each n ∈ N, let P (n) be the statement “Tn < 2n ”.
Base step. Clearly P (1), P (2), P (3) are true.
Induction step. Let k ≥ 3 and Assume P (1), . . . , P (k) are all true. Now
Tk+1 = Tk + Tk−1 + Tk−2 < 2k + 2k−1 + 2k−2 < 2k+1 .
So P (k + 1) is true.
Hence, by induction, P (n) is true for all n ∈ N.
12. The inductive step is not valid because the set {a − 1, b − 1} might no longer
be a subset of N.
13.
(i) S1 (n) = n(n + 1)/2, S2 (n) = n(n + 1)(2n + 1)/6 and S3 (n) = n2 (n +
1)2 /4.
(ii) Substitute x = 1, 2, . . . n in (x + 1)5 − x5 = 5x4 + 10x3 + 10x2 + 5x + 1:
25 − 15 = 5 · 14 + 10 · 13 + 10 · 12 + 5 · 1 + 1,
35 − 25 = 5 · 24 + 10 · 23 + 10 · 22 + 5 · 2 + 1,
..
.. ..
..
.
. .
.
n5 − (n − 1)5 = 5 · (n − 1)4 + 10 · (n − 1)3 + 10 · (n − 1)2 + 5 · (n − 1) + 1,
(n + 1)5 − n5 = 5 · n4 + 10 · n3 + 10 · n2 + 5 · n + 1.
Adding up we get (n + 1)5 − 1 = 5S4 (n) + 10S3 (n) + 10S2 (n) + 5S1 (n) +
S0 (n).
(iii) Substitute the expressions for S1 , S2 and S3 from part (i) and simplify.
(Note S0 (n) = n.)
3n2 + 3n − 1
is an integer.
(iv) 12 + 22 + . . . + n2 divides 14 + 24 + . . . + n4 iff
5
Writing n = 5m + k where m ∈ Z and k = 0, 1, 2, 3, 4 we see that
3n2 + 3n − 1
is an integer iff k = 1 or k = 3.
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14.
(i) Pair up the contestants. Contestants in a pair stand close eg at arms
length, and the pairs are placed quite apart (eg 10 metres apart).
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(ii) Measure distances between pairs of contestants and pick a pair, say A and
B, with distance minimum among all pairings. Then A must be the unique
nearest neighbour of B, for otherwise there is a contestant C nearer to A
than B is to A and that contradicts our choice of A, B having minimal
distance amongst pairs of contestants. Similary B is the unique nearest
neighbour of A, and so A, B throw tomatoes at each other.
(iii) We prove the result by induction on the number of contestants n = 2k + 1
with k = 0, 1, 2, 3, . . ..
If k = 0 i.e. there is just one contestant, then there is obviously a winner.
For the inductive step, assume that a competition with n = 2k + 1 contestants always produces a winner/winners. Now consider a competition
with exactly 2k + 3 contestants. By the previous part, we can pick a pair
of contestants A and B who throw tomatoes at each other. There are
now two cases to consider.
First, suppose no contestant other than A, B throw their tomato to A or B.
Then the 2k + 1 contestants different from A and B throw tomatoes only
amongst themselves. By the inductive hypothesis, there is a contestant
amongst this 2k + 1 players, say C, who doesnt get hit by a tomato from
the other 2k players. By our choice neither A nor B throws tomatoes at
C. So C will be a winner.
Let’s now consider the case when someone from the 2k + 1 contestants
different from A and B direct their tomatoes to A or B. Now 2k + 3
tomatoes are thrown altogether with 2k + 3 possible targets. Since A and
B together gets hit by at least 3 tomatoes, there arent enough tomatoes
to hit the remaining 2k + 1, and therefore there will be at least one winner.
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