Sigma Notation
Introduction
The sigma notation is a shorthand for writing sums. In this lesson, we will
see the power of this notation in computing sums of numbers as well as algebraic
expressions.
2.2.1. Writing and Evaluating Sums in Sigma Notation
Mathematicians use the sigma notation to denote a sum. The uppercase Greek
letter ⌃ (sigma) is used to indicate a “sum.” The notation consists of several
components or parts.
Let f (i) be an expression involving an integer i. The
expression
f (m)+ f (m + 1)+ f (m + 2)+ ·· · + f (n)
can be compactly written inn sigma notation, and we
write it as X
f
i=m (i),
which is read “the summation of f (i) from i = m to n.” Here, m
and n are integers with m n, f (i) is a term (or summand )
of the summation, and the letter i is the index, m the lower
bound, and n the upper bound.
Example 2.2.1. Expand each summation, and simplify if possible.
Xn
X4
(1) (3) ai
(2i +
3) i=1
i=2
X6 p n
X5 n
(2) i (4)
2 +1
i=0 n=
1
Solution. We apply the definition of sigma notation.
X
4
(1) (2i + 3) = [2(2)+ 3]+ [2(3)+ 3]+ [2(4)+ 3] = 27
i=2
5
(2) X
2i = 20 + 21 + 22 + 23 + 24 + 25 = 63
i=
(3) X
0
n
ai = a1 + a2 + a3 + ·· · + an
i=1
p p p p
X p 1 2 3 2 5 6
n = + + + + 2
(4) n +1 +2 3
4 5 6 7
6
n=
1
Example 2.2.2. Write each expression in sigma notation.
1 1 1 1
(1) 1+ + + + ·· · +
2 3 4 100
(2) — 1 + 2 — 3 + 4 — 5 + 6 — 7 + 8 — 9 + ·· · — 25
(3) a2 + a4 + a6 + a8 + ·· · + a20
1 1 1 1 1 1
(4) 1 + + 1
+ + +
2 4 + +
8 16 32 64 128
1 1 1
100 X
1
Solution. (1) 1+ + + + ·· · + = 1 n
2 3 4 100 n=1
(2) — 1 + 2 — 3 + 4 — 5 + ·· · — 25
= (—1)1 1 + (—1)2 2 + (—1)3 3 + (—1)4 4
+ (—1)5 5 + ·· · + (—1)25 25
X25
= (—1)jj
j=1
(3) a2 + a4 + a6 + a8 + ·· · + a20
= a2(1) + a2(2) + a2(3) + a2(4) + ·· · + a2(10)
X
10
= a2i
i=1
7
1 1 1
1 1 1 1 X 1
k=0
(4) 1 + + + + + + + = 2
2 4 8 16 32 64 128 2k
The sigma notation of a sum expression is not necessarily unique. For ex-
ample, the last item in the preceding example can also be expressed in sigma
notation as follows:
1 1 1 1 X
8
1
1 1 1
k=1
1+ + + + + + + =
2 4 8 16 2k—1
32 64 128
.
However, this last sigma notation is equivalent to the one given in the example.
Seatwork/Homework 2.2.1
1. Expand each summation, and simplify if possible.
(a) X
5
(2 — 3k) Answer: —28
k=—
(b) 1
Xn
(c) xj Answer: x + x2 + x3 + ·· · + xn
j=
X
(d) 1
6
(e) (j2 — j) Answer: 68
j=
X
3
4
(—1)k+1k Answer: —2
k=
X
1
3
(an+1 — an) Answer: a4 — a1
n=
1
2. Write each expression in sigma notation.
(a) x + 2x2 + 3x3 + 4x4 + 5x5 5
Answer: X
kxk
X
10 k=
1
(b) 1 — 2 + 3 — 4 + 5 — 6 + ·· · — 10 Answer: (—1)k+1 k
k=1
(c) 1 + 3 + 5 + 7 + ·· · + 101 Answer: 50
X
(2k + 1)
k=0
X 4
(d) a4 + a8 + a12 + a16 Answer: a4k
4 k
k=
1
1 1 1 X (—1)
1 (e) 1 —
+ — + Answer:
3 5 7 9 2k +1
k=0
2.2.2. Properties of Sigma Notation
We start with finding a formula for the sum of
n
X
i = 1 + 2 + 3 + ·· · + n
i=1
in terms of n.
The sum can be evaluated in di↵erent ways. A simple, though informal,
approach is pictorial.
T
to guess.
n
i X = 1 + 2 + 3 + + n = n(n +
1i=1
) 2
·· ·
Another way is to use the formula for an arithmetic series with a1 = 1 and
an = n:
n(a1 + an) n(n + 1)
S= 2 = 2 .
We now derive some useful summation facts. They are based on the axioms
of arithmetic addition and multiplication.
Xn Xn
cf (i) = cf (i), c any real
i=m i=m number.
Proof. n
X
cf (i) = cf (m)+ cf (m + 1)+ cf (m + 2)+ ·· · + cf
(n)
i=
m
= c[f (m)+ f (m + 1)+ ·· · + f (n)]
Xn
=c f (i) 2
i=m
Xn Xn Xn
[f (i)+ g(i)] =f (i)+g(i)
i=m i=m i=m
Proof.
n
X
[f (i)+ g(i)]
i=m
= [f (m)+ g(m)] + ·· · + [f (n)+ g(n)]
= [fn(m)+ ·· ·n+ f (n)] + [g(m)+ ·· · + g(n)]
X X
= f (i)+ g(i) 2
i=m i=m
Xn
c = c(n — m +
i=m 1)
Proof.
X
n
c = c + c + c + ·· ·
| {z
+c }
i=m
n—m+1 terms
= c(n — m + 1) 2
A special case of the above result which you might encounter more often is
the following:
Xn
c = cn.
i=1
Telescoping
Sum
Xn
[f (i + 1) — f (i)] = f (n + 1) —
i=m f (m)
Proof.
X
n ⇥ ⇤
f (i + 1) — f (i)
i=m
= [f (m + 1) — f (m)] + [f (m + 2) — f (m + 1)]
+ [f (m + 3) — f (m + 2)]+ ·· · + [f (n + 1) — f (n)]
Note that the terms, f (m + 1),f (m + 2),. . . , f (n), all cancel out. Hence, we have
n
X
[f (i + 1) — f (i)] = f (n + 1) — f (m). 2
i=m
Example 2.2.3. Evaluate:
30
X
(4i — 5).
i=1
Solution.
30 30 30
X X X
(4i — 5) = 4i — 5
i=1 i=1 i=
30 1
3
0
X X
=4 i— 5
(30)(31)i=1
i=1
=4 — 5(30)
2
= 1710 2
Example 2.2.4. Evaluate:
1 1 1 1
+ + + ·· · + .
1·2 2·3 3·4 99 · 100
Solution.
1 1 1 1
+ + + ·· · +
1·2 2·3 3·4 99 · 100
X
99
1
= i(i + 1)
i=1
X99 i + 1
=— i
i(i + 1)
i=1
Σ
X
99 i+1 i
=
—
i(i + 1) i(i +
i=1 ◆
99 ✓ 1) 1
X 1
= —
i i +1
i=1 ✓ ◆
1 1
X9
9
=— —
i +1 i
i=1
1
Using f (i) = and the telescoping-sum property, we get
i
99 ✓
1 ◆
X 1 1 99
= — = . 2
i=1 i(i + 1) 100 100
—
1
n
X
Example 2.2.5. Derive a formula for i2 using a telescoping sum with terms
i=1
f (i) = i3.
Solution. The telescoping sum property implies that
X
n ⇥ ⇤
i — (i — 1)3 = n3 — 03 = n3.
3
i=1
On the other hand, using expansion and the other properties of
summation, we have
n n
X⇥ ⇤ X
i — (i — 1) =
3 3
(i3 — i3 + 3i2 — 3i + 1)
i=1 i=1
n n n
X X X
= 3i=
1n i 2
— 3
i=1 i + i=11
X n(n + 1)
=3 i2 — 3 · + n.
2
i=1
Equating the two results above, we obtain
X n 3n(n +
3 i2 + n = n3
1) 2
—
i=1
X n
6 i2 — 3n(n + 1)+ 2n = 2n3
i=1
n
X
6 i2 = 2n3 — 2n + 3n(n + 1)
i=1
= 2n(n2 — 1) + 3n(n + 1)
= 2n(n — 1)(n + 1)+ 3n(n + 1)
= n(n + 1)[2(n — 1) + 3]
= n(n + 1)(2n + 1).
Finally, after dividing both sides of the equation by 6, we obtain the
desired formula n
X
i2 = n(n + 1)(2n + . 2
1) 6
i=1
Seatwork/Homework 2.2.2
1. Use the properties of sigma notation to evaluate the following summations.
(a)
50
X
(2 — 3k) Answer: —3725
(b)
k=
X1
n
(1 + 2j) Answer: 2n + n2
j=1
99
X
(c) 1
Answer: 9
j= i + 1 +
p
p i
1
Solution:
99
99 p
X
X 1 p= p 1 i +1 — i
p p·p p
j=1 i + 1 + i i +1+ i i +1 — i
p⌘
j=1 i +1 i
X
9 ⇣ —
9 p
=
j=
1
p p
= 99 + 1 — 1
=9
n
X
2. If (i + 1)2 = an3 + bn2 + cn + d, what is a + b + c + d? Answer: 4
i=1
Exercises 2.2
1. Expand each sum.
X
9 i 1 2 3 4 5
(a) Answer: + + + +
i= x + x +1 x +2 x +3 x +4 x +5
5 i
X
6
p3 p3 p3 p3 p3 p3
(b) 2i Answer: 0 + 2+ 4+ 6+2+ 10 + 12
i=
X
0
3
(c) 3—i Answer: 9 + 3 + 1 + 1/3+ 1/9+ 1/27
i=—2
2. Write each expression in sigma notation.
12
(a) 1 + 22 + 33 + 44 + ·· · + 1212 Answer: X i
i
(b) (x — 5) + (x — 3) + (x — 1) + (x + 1)+ (x + 3)+ ·· · + (x + 15) i=1
7 Another
possible
answer for
(b) is
X X11 [x + (2i —
Answer: [x + (2i + 1)]
i=17)].
i=—3
X
(c) a1 + a4 + a9 + a16 + ·· · + a81 Answer: 9
ai
3. Evaluate each sum. 2
i=1
(a) X
120
(4i — 15) Answer: 27240
i=
(b) X
1
5 [(5i — 2)(i + 3)] Answer: 230900
i=1
0 n 3 2
X 6n + 3n — 3n + 2
(c) (3i — 1)2 Answer:
2
i=1
30
X 30
X 30
X 3g(i) — f(i)+7
4. If f (i) = 70 and g(i) = 50, what is the value of ?
i=1 2
i=1 i=1
Answer: 145
100 200
X X
5. If s = i, express i in terms of s. Answer: 2s + 100000
i= i=1
1 n
n i
X X
6. If s =i=1 ai, does it follow that i=1a2 = s2?
2 n
Answer: No. If s = X X
ai = a1 + a2, then a2 = a2 + a2, while s2 =
i 1 2
a2 + 2a1a2 + a2. i=1 i=1
1 2
Xn
7. Derive a formula for i3 by using a telescoping sum with terms f (i) = i4.
i=1
Answer: n2(n + 1)2
4
