54 Chapter 4
Let us sum now the geometric series S = a + ar + ar
2
+ ··· + ar
n−1
. Plainly, if r = 1 then S = na, so we may assume that r
6= 1. We have rS = ar + ar
2
+ ··· + ar
n
. Hence
S − rS = a + ar + ar
2
+ ··· + ar
n−1
− ar − ar
2
− ··· − ar
n
= a − ar
n
. From this we deduce that
S = a − ar
n
1 − r ,
that is, a + ar +
···+ ar
n−1
= a − ar
n
1 − r 4.3
If |r| 1 then r
n
→ 0 as n → ∞
. For
|r| 1, we obtain the sum of the infinite geometric series a + ar + ar
2
+ ··· =
a 1 − r
4.4
368 Example A fly starts at the origin and goes 1 unit up, 1
2 unit right, 14 unit down, 18 unit left, 116 unit up, etc., ad infinitum. In what coordinates does it end up?
Solution: Its x coordinate is 1
2 −
1 8
+ 1
32 −
··· =
1 2
1 −
−1 4
= 2
5 .
Its y coordinate is 1 −
1 4
+ 1
16 −
··· = 1
1 −
−1 4
= 4
5 .
Therefore, the fly ends up in 2
5 ,
4 5
.
Practice
369 Problem The 6th term of a geometric progression is 20 and the 10th is 320. Find
i its 15th term, ii the sum of its first 30 terms.
4.4 Fundamental Sums
In this section we compute several sums using telescoping cancellation. We start with the sum of the first n positive integers, which we have already computed using Gauss’ trick.
370 Example Find a closed formula for
A
n
= 1 + 2 + ···+ n.
Solution: Observe that k
2
− k − 1
2
= 2k − 1 .
Fundamental Sums 55
From this 1
2
− 0
2
= 2
· 1 − 1 2
2
− 1
2
= 2
· 2 − 1 3
2
− 2
2
= 2
· 3 − 1 ..
. ..
. ..
. n
2
− n − 1
2
= 2
· n − 1 Adding both columns,
n
2
− 0
2
= 21 + 2 + 3 + ···+ n − n.
Solving for the sum, 1 + 2 + 3 +
···+ n = n
2
2 + n2 = nn + 1
2 .
371 Example
Find the sum 1
2
+ 2
2
+ 3
2
+ ··· + n
2
. Solution: Observe that
k
3
− k − 1
3
= 3k
2
− 3k + 1 .
Hence 1
3
− 0
3
= 3
· 1
2
− 3 · 1 + 1
2
3
− 1
3
= 3
· 2
2
− 3 · 2 + 1
3
3
− 2
3
= 3
· 3
2
− 3 · 3 + 1
.. .
.. .
.. .
n
3
− n − 1
3
= 3
· n
2
− 3 · n + 1
Adding both columns, n
3
− 0
3
= 31
2
+ 2
2
+ 3
2
+ ··· + n
2
− 31 + 2 + 3 + ···+ n + n.
From the preceding example 1 + 2 + 3 + ···+ n = ·n
2
2 + n2 = nn + 1
2 so
n
3
− 0
3
= 31
2
+ 2
2
+ 3
2
+ ··· + n
2
− 3
2 · nn + 1 + n.
Solving for the sum, 1
2
+ 2
2
+ 3
2
+ ··· + n
2
= n
3
3 +
1 2
· nn + 1 − n
3 .
After simplifying we obtain 1
2
+ 2
2
+ 3
2
+ ··· + n
2
= nn + 12n + 1
6 4.5
372 Example
Add the series 1
1 · 2
+ 1
2 · 3
+ 1
3 · 4
+ ··· +
1 99
· 100 .
56 Chapter 4
Solution: Observe that 1
kk + 1 =
1 k
− 1
k + 1 .
Thus 1
1 · 2
= 1
1 −
1 2
1 2
· 3 =
1 2
− 1
3 1
3 · 4
= 1
3 −
1 4
.. .
.. .
.. .
1 99
· 100 =
1 99
− 1
100 Adding both columns,
1 1
· 2 +
1 2
· 3 +
1 3
· 4 +
··· + 1
99 · 100
= 1 − 1
100 =
99 100
.
373 Example Add
1 1
· 4 +
1 4
· 7 +
1 7
· 10 +
··· + 1
31 · 34
. Solution: Observe that
1 3n + 1
· 3n + 4 =
1 3
· 1
3n + 1 −
1 3
· 1
3n + 4 .
Thus 1
1 · 4
= 1
3 −
1 12
1 4
· 7 =
1 12
− 1
21 1
7 · 10
= 1
21 −
1 30
1 10
· 13 =
1 30
− 1
39 ..
. ..
. ..
. 1
34 · 37
= 1
102 −
1 111
Summing both columns, 1
1 · 4
+ 1
4 · 7
+ 1
7 · 10
+ ··· +
1 31
· 34 =
1 3
− 1
111 =
12 37
.
374 Example Sum
1 1
· 4 · 7 +
1 4
· 7 · 10 +
1 7
· 10 · 13 +
··· + 1
25 · 28 · 31
. Solution: Observe that
1 3n + 1
· 3n + 4 · 3n + 7 =
1 6
· 1
3n + 13n + 4 −
1 6
· 1
3n + 43n + 7 .
Practice 57
Therefore 1
1 · 4 · 7
= 1
6 · 1 · 4
− 1
6 · 4 · 7
1 4
· 7 · 10 =
1 6
· 4 · 7 −
1 6
· 7 · 10 1
7 · 10 · 13
= 1
6 · 7 · 10
− 1
6 · 10 · 13
.. .
.. .
.. .
1 25
· 28 · 31 =
1 6
· 25 · 28 −
1 6
· 28 · 31 Adding each column,
1 1
· 4 · 7 +
1 4
· 7 · 10 +
1 7
· 10 · 13 +
··· + 1
25 · 28 · 31
= 1
6 · 1 · 4
− 1
6 · 28 · 31
= 9
217 .
375 Example Find the sum
1 · 2 + 2 · 3 + 3 · 4 + ···+ 99 · 100.
Solution: Observe that kk + 1 =
1 3
kk + 1k + 2 − 1
3 k − 1kk + 1
. Therefore
1 · 2
= 1
3 · 1 · 2 · 3 −
1 3
· 0 · 1 · 2 2
· 3 =
1 3
· 2 · 3 · 4 − 1
3 · 1 · 2 · 3
3 · 4
= 1
3 · 3 · 4 · 5 −
1 3
· 2 · 3 · 4 ..
. ..
. ..
. 99
· 100 = 1
3 · 99 · 100 · 101 −
1 3
· 98 · 99 · 100 Adding each column,
1 · 2 + 2 · 3 + 3 · 4 + ···+ 99 · 100 =
1 3
· 99 · 100 · 101 − 1
3 · 0 · 1 · 2 = 333300.
Practice
376 Problem Shew that
1
3
+ 2
3
+ 3
3
+ ··· + n
3
=
nn + 1 2
2
4.6
377 Problem Let a
1
, a
2
, . . . ,a
n
be arbitrary numbers. Shew that a
1
+ a
2
1 + a
1
+ a
3
1 + a
1
1 + a
2
+a
4
1 + a
1
1 + a
2
1 + a
3
+ ···
+a
n−1
1 + a
1
1 + a
2
1 + a
3
··· 1 + a
n−2
= 1 + a
1
1 + a
2
1 + a
3
··· 1 + a
n
− 1 .
378 Problem Shew that
csc 2 + csc 4 + csc8 + ··· + csc2
n
= cot 1 − cot 2
n
.
379 Problem Let 0
x 1. Shew that
∞
X
n=1
x
2n
1 − x
2n+1
= x
1 − x .
58 Chapter 4
380 Problem
Shew that tan
π
2
100
+ 2 tan
π
2
99
+2
2
tan
π
2
298
+ ··· + 2
98
tan
π
2
2
= cot
π
2
100
.
381 Problem Shew that
n
X
k=1
k k
4
+ k
2
+ 1 =
1 2
· n
2
+ n n
2
+ n + 1 .
382 Problem Evaluate
1 · 2 · 4 + 2 · 4 · 8 + 3 · 6 · 12 + ···
1 · 3 · 9 + 2 · 6 · 18 + 3 · 9 · 27 + ···
1 3
.
383 Problem Shew that
∞
X
n=1
arctan 1
1 + n + n
2
=
π
4 .
Hint: From tan x − tan y =
tan x − tan y 1 + tan x tan y
deduce that arctan a − arctan b = arctan
a − b 1 + ab
for suitable a and b.
384 Problem
Prove the following result due to Gramm
∞
Y
n=2
n
3
− 1 n
3
+ 1 =
2 3
.
4.5 First Order Recursions
