Chapter
4
Sums, Products, and Recursions
4.1 Telescopic cancellation
We could sum the series
a
1
+ a
2
+ a
3
+ ··· + a
n
if we were able to find {v
k
} satisfying a
k
= v
k
− v
k−1
. For a
1
+ a
2
+ a
3
+ ··· + a
n
= v
1
− v + v
2
− v
1
+ ··· + v
n−1
− v
n−2
+ v
n
− v
n−1
= v
n
− v .
If such sequence v
n
exists, we say that a
1
+ a
2
+ ··· + a
n
is a telescopic series.
341 Example Simplify
1 + 1
2
·
1 + 1
3
·
1 + 1
4
···
1 + 1
99
.
Solution: Adding each fraction: 3
2 ·
4 3
· 5
4 ···
100 99
, which simplifies to 100
2 = 50.
342 Example Find integers a
, b so that 2 + 1
·
2
2
+ 1
·
2
2
2
+ 1
·
2
2
3
+ 1
···
2
2
99
+ 1
= 2
a
+ b .
47
48 Chapter 4
Solution: Using the identity x
2
− y
2
= x − yx + y and letting P be the sought product: 2 − 1P
= 2 − 1 2 + 1
·
2
2
+ 1
·
2
2
2
+ 1
·
2
2
3
+ 1
···
2
2
99
+ 1
=
2
2
− 1
·
2
2
+ 1
·
2
2
2
+ 1
·
2
2
3
+ 1
···
2
2
99
+ 1
=
2
2
2
− 1
·
2
2
2
+ 1
·
2
2
3
+ 1
···
2
2
99
+ 1
=
2
2
3
− 1
·
2
2
3
+ 1
·
2
2
4
+ 1
···
2
2
99
+ 1
.. .
.. .
= 2
2
99
− 12
2
99
+ 1 =
2
2
100
− 1 ,
whence P = 2
2
100
− 1 .
343 Example Find the exact value of the product
P = cos
π
7 · cos
2
π
7 · cos
4
π
7 .
Solution: Multiplying both sides by sin
π
7 and using sin 2x = 2 sin x cos x we obtain
sin
π
7 P
= sin
π
7 cos
π
7 · cos
2
π
7 · cos
4
π
7 =
1 2
sin 2
π
7 cos
2
π
7 · cos
4
π
7 =
1 4
sin 4
π
7 cos
4
π
7 =
1 8
sin 8
π
7 .
As sin
π
7 = − sin
8
π
7 , we deduce that
P = − 1
8 .
344 Example Shew that
1 2
· 3
4 ·
5 6
··· 9999
10000 1
100 .
Solution: Let A =
1 2
· 3
4 ·
5 6
··· 9999
10000 and
B = 2
3 ·
4 5
· 6
7 ···
10000 10001
. Clearly, x
2
− 1 x
2
for all real numbers x. This implies that x − 1
x x
x + 1
Practice 49
whenever these four quantities are positive. Hence 1
2 2
3 3
4 4
5 5
6 6
7 ..
. ..
. ..
. 9999
10000 10000
10001 As all the numbers involved are positive, we multiply both columns to obtain
1 2
· 3
4 ·
5 6
··· 9999
10000 2
3 ·
4 5
· 6
7 ···
10000 10001
, or A
B. This yields A
2
= A · A A · B. Now
A · B =
1 2
· 2
3 ·
3 4
· 4
5 ·
5 6
· 6
7 ·
7 8
··· 9999
10000 ·
10000 10001
= 1
10001 ,
and consequently, A
2
A · B = 110001. We deduce that A 1 √
10001 1100.
For the next example we recall that n n factorial means n = 1
· 2 · 3···n. For example, 1 = 1
, 2 = 1 · 2 = 2, 3 = 1 · 2 · 3 = 6,4 = 1 · 2 · 3 · 4 = 24. Observe that k + 1 = k + 1k. We make the convention 0 = 1
.
345 Example
Sum 1
· 1 + 2 · 2 + 3 · 3 + ···+ 99 · 99. Solution: From k + 1 = k + 1k = k
· k + k we deduce k + 1 − k = k · k. Thus 1
· 1 =
2 − 1 2
· 2 =
3 − 2 3
· 3 =
4 − 3 ..
. ..
. ..
. 98
· 98 =
99 − 98 99
· 99 = 100 − 99 Adding both columns,
1 · 1 + 2 · 2 + 3 · 3 + ···+ 99 · 99 = 100 − 1 = 100 − 1.
Practice
50 Chapter 4
346 Problem
Find a closed formula for D
n
= 1 − 2 + 3 − 4 + ··· + −1
n−1
n .
347 Problem
Simplify
1 − 1
2
2
·
1 − 1
3
2
·
1 − 1
4
2
···
1 − 1
99
2
.
348 Problem
Simplify log
2
1 + 1
2
+ log
2
1 + 1
3
+ log
2
1 + 1
4
+ ··· + log
2
1 + 1
1023
.
349 Problem Prove that for all positive integers n, 2
2n
+ 1 divides 2
22 n
+1
− 2 .
4.2 Arithmetic Sums
Kategori