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MAT Sequences Practice
1. Find the sum of the first 𝑛 terms
for the following sequences:
a. 1, 2, 4, 8, ...
1 1 1
2 4 8
1 1
1
1, − 2 , 4 , − 8 , …
b. 1, , , , ...
c.
1
1
a. 9, 9, 3, 27, 1, 81, 3 , 243, …
b. 1, 1, 2, 3, 4, 9, 8, 27, 16, 81, …
1
4
1
8
c. 1, 1, , 2, , 3, , 4,
1
, 5, …
16
3. Prove that the sum of the first 𝑛
1
natural numbers is 2 𝑛(𝑛 + 1).
4. Determine the sum to infinity to
the following sequences:
1
1
a. 2, −1, 2 , − 4 , …
b.
1 1 1 1 1 1
3, 2, 1, 1, , , , , , , …
3 2 9 4 27 8
5. Determine the sum of the
following arithmetic sequences:
a. 1 + 2 + 3 + ⋯ + 200
b. 1 + 3 + 5 + ⋯ + 99
c. 2 + 4 + 6 + ⋯ + 100
6. Given a sequence 𝑥𝑛 = 𝑛2 − 2𝑛,
determine the increase of the 𝑛th
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1
2
3
5
infinity of 1 + 2 + 4 + 8 + 16 +
d. 1, −1, 1, −1, …
2. Find the sum of the first 2𝑛 terms
of the following sequences (note
that there are two sequences
interleaved!):
1
2
term from the previous term. For
what value of 𝑛 will the increase
from the previous term by 99?
7. [Source: SMC] Find the sum to
8
32
13
+ 64 + ⋯ (Hint: Since the
numerators have the property
that they’re the sum of the
previous two numerators,
perhaps we could replace all but
the first few fractions each with a
pair of fractions added together?)
8. [Source: MAT] Given that 1 +
1
22
1
1
+ 32 + 4 2 + ⋯ =
1+
1
32
+
1
52
+
1
72
𝜋2
6
and that
+⋯=
𝜋2
,
8
determine the value of the infinite
1
1
1
1
1
sum 1 − 22 + 32 − 42 + 52 − 62 +
⋯
ANSWERS
1.
a. 𝑆𝑛 = 2𝑛−1
b. 𝑆𝑛 = 2 − 21−𝑛
2
2
c. 𝑆𝑛 = 3 − 3 (−2)−𝑛
d. 𝑆𝑛 = (−1)𝑛−1
2.
27
1
(3−𝑛 ) + 3𝑛+2
2
2
𝑛−1
𝑛−1
a. 9 −
b. 2
+3
1−𝑛
c. 2
1
+ 2 𝑛(𝑛 + 1)
3. Just pick your favourite proof! You could observe that the first and last term add up to 𝑛 +
𝑛
2
1, as does the next pair inwards, and so on. There’s of these pairs, so the total is
1
𝑛(𝑛
2
+ 1). Alternatively, observe that the median of the numbers is
𝑛+1
2
by observation,
and that the mean is the equal to the median due to symmetry. There’s 𝑛 numbers, so the
1
total is 2 𝑛(𝑛 + 1).
4.
4
3
17
2
a.
b.
5. Determine the sum of the following arithmetic sequences:
1
a. 1 + 2 + 3 + ⋯ + 200: The sum of 1 to n is 2 𝑛(𝑛 + 1), which gives 20100.
b. 1 + 3 + 5 + ⋯ + 99: 𝑎 = 1 and 𝑑 = 2 and 𝑛 = 50 (Be very careful when working
out the number of terms. Notice that if we added 1 to all the numbers then divided
by 2, we’d have the numbers 1 to 50, and thus it then becomes clear there are 50
numbers). Thus 𝑆50 =
50
(2 +
2
49 × 2) = 2500
c. 2 + 4 + 6 + ⋯ + 100: 𝑎 = 2, 𝑑 = 2, 𝑛 = 50. So 𝑆50 =
50
(4
2
+ 49 × 2) = 2550
6. 𝑥𝑛 − 𝑥𝑛−1 = 𝑛2 + 2𝑛 − (𝑛 − 1)2 − 2(𝑛 − 1)
= 2𝑛 + 1
Then if 2𝑛 + 1 = 99, 𝑛 = 49.
7.
1
1
2
3
5
8
13
1
1
1
1
2
1 1 2 3
5
1 1 2
3
= 1+( + + +
+
+ ⋯) +( + +
+ )
2 4 8 16 32
4 8 16 32
1
1
= 1 + 𝑆∞ + 𝑆∞
2
4
1
𝑆 = 1, so 𝑆∞ = 4.
4 ∞
8.
2
3
𝑆∞ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + ⋯ = 1 + (2) + (4 + 4) + (8 + 8) + (16 + 16) + ⋯
1
22
1
1
+ 4 2 + 62 + ⋯ =
1
1
Then 1 − 22 + 32 −
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𝜋2
𝜋2
𝜋2
−
=
6
8
24
1
1
𝜋2
+ 52 + ⋯ = 8
42
𝜋2
𝜋2
− 24 = 12