EECS 210
Supplementary notes on summations
These notes discuss some general methods that can be applied to summations and contain
some formulas that frequently occur. Let's begin by looking at the sum of the terms of a
sequence in which all terms are the same value. For example,
n
n
if each ai = 1, then ai =  1 = a1 + a2 + ... an = 1 + 1 + ... + 1 = n
i=1
i=1
In general, if each ai = c where c is a constant, we have:
n
n
ai =  c = a1 + a2 + ... an = c + c + ... + c = n•c
i=1
i=1
n
Similarly,  j = n•j since the index of summation is i and thus j is treated as a constant.
i=1
Methods of manipulating sums and using formulas
When evaluating complicated sums, there are several algebraic techniques that often enable us
to break our problem up into parts which can be handled separately. Let's begin with the
following formulas:
n
n
1) cai = c ai
i=1
i=1
n
n
n
2)(ai + bi) = ai +  bi
iii
nn-1
3)  (ai - ai-1) = an - a0 and  (ai - ai+1) = a0 - an
i=1
n
i=0
n+k
n
n
m-1
4)  ai = ai-k and  ai =  ai -  ai
i=m
i=m+k
i=m
i=1
From the text we have:
i=1
n
5) ari = (arn+1 - a)/(r - 1) if r  1
(Geometric Series)
i=0
n
6) i = (n)(n+1)/2
i=1
n
7) i2 = (n)(n+1)(2n+1)/6
i=1
We begin with the following sequences:
1, 3, 5, 7, ..., 2n + 1
and
2, 6, 10, 14, ..., 4n – 2
In both sequences, each term is obtained from its predecessor by adding a constant--two in the
first case, and four in the second. These are examples of arithmetic sequences and could be
written in summation notation as follows:
n
n+1
 (2i + 1) and  (4i - 2), respectively.
i=0
i=1
n
n
n
Using 2) and 6) above, (2i +1) = 2 i + 1 = 2(n)(n + 1)/2 + (n + 1) = n2 + 2n + 1 = (n+1) 2
i=0
i=0
i=0
There are a couple of things to note here. First of all, observe that each sum has n + 1 terms
since the index of summation starts at 0. Second, we are able to break the sum apart into two
parts and evaluate each separately. The first sum is obtained by using 1) and 6). Since the first
term is 0, 6) can be used without modification. In general, an arithmetic sequence has the
following form: a, (a + d), (a + 2d), ... , (a + n·d) and the corresponding sum is:
n
n
n
n
n
[a + (a + d) + ... + (a + n•d)] =  (a + i•d) = a +  i•d = a + d•i = (n + 1)a +d•(n)(n + 1)/2
i=0
i=0
i=0
i=0
i=0
Suppose you are asked to find the sum of the integers from 50 to 100, inclusive. This can be
done quite easily using formula 6) and some algebra:
100
100
49
 i =  i -  i = (100)(101)/2 - (49)(50)/2 = 5050 - 1225 = 3825
i=50
i=1
i=1
Consider the following pseudocode:
x0
for i  1 to n do
for j  i to n do
xx+1
What is the value of x when the block of code terminates? When i = 1, j goes from 1 to n and
adds n to the value of x. When i = 2, j goes from 2 to n and adds n - 1 to the value of x. Then,
the value of x when the loop terminates can be found by computing the following sum:
n
n + (n - 1) + (n - 2) + ... + 1 or  i = (n)(n+1)/2 by formula 6) above.
i=1
Now, let's consider an example of using one of the above formulas to find a formula for a
different sum. Consider the first arithmetic sequence above, namely, the sum of consecutive odd
numbers. As expressed above, indexing from 0 to n, gives us the sum of the first n+1 odd
numbers. Using 2n -1 to represent an odd number and formula 6) above, we can obtain a
formula for the sum of the first n odd numbers as follows:
n
n
n
1 + 3 + ... + 2n - 1 = (2i - 1) = 2  i -  1 = 2(n)(n+1)/2 - n = n2 + n - n = n2
i=1
i=1
i=1
As a more complex example of deriving one formula from another, let's use 3) above to obtain
formula 7)
n
n3
=
n3
- 0 =  [i3 - (i – 1)3] Using 2) and 1) we then get
i=1
n
n
n
n
n
n
n
n
n
n3 =  i3 -  (i – 1)3 =  i3 - ( i3 - 3 i2 + 3i - 1) =   i3 -  i3 + 3  i2 - 3 i +  1
i=1
i=1
i=1
i=1
i=1
i=1
i=1
i=1
i=1
n
So,
n3
= 3  i2 - 3 (n)(n+1)/2 +n
i=1
n
Then, 3  i2 n3 + 3(n)(n+1)/2 + n = n(n2 +3/2 n + 1/2) = (n/2)(2n2 + 3n + 1)
i=1
n
and  i2 = (n/6)(2n2 + 3n + 1) = (n/6)(n + 1)(2n + 1)
i=1
Other solution methods
Let's begin by looking at one way of finding the sum of a geometric series that doesn't rely on
formula 5) above. Although we'll look at a specific example, formula 5) can be derived in the
same way. A sum that frequently arises when we study binary trees is the geometric series:
1 + 2 + 4 + ... + 2n. Let S represent this sum.
Multiplying by 2 we have 2S = 2 + 4 + ... +2n+1 Then, finding the difference,
2S =
2 + 4 + ... + 2n + 2n+1
-S = -1 - 2 - 4 - ... - 2n
_________________________
S = -1 + 2n+1
Notice that most of the terms cancelled out leaving us with S = 2n+1 - 1. Solving sums by
telescoping is done in much the same way, namely, expanding the summation permits us to
cancel several terms and obtain a concise closed form solution. Consider the following sum, and
observe what happens when we group successive terms
n
[ k/(k+1) - (k+1)/(k+2)] = (1/2 - 2/3) + (2/3 - 3/4) + (3/4 - 4/5) + ... [(n/(n+1) - (n+1)/(n+2)]
k=1
= 1/2 + (- 2/3 + 2/3) + (- 3/4 + 3/4) + ... + [-n/(n+1) +n/(n+1)] - (n+1)/(n+2)
= 1/2 - (n+1)/(n+2)
This same approach can be employed to find sums like the following:
1/ (1•2) + 1/ (2•3) + 1/ (3•4) + ... + 1/ [n(n+1)].
Observe that the nth term can be rewritten as follows: 1/ [n(n+1)] = 1/n - 1/(n+1). Then, using
telescoping we obtain the closed form solution below.
n
[1/k - 1/(k+1)] = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ...+ [(1/n - 1/(n+1)] = 1 - 1/(n+1)
k=1
n
As a final example of this method, let's find the closed form solution of  i·2i. Letting
i=1
S be the sum, writing the terms in reverse order and finding the difference 2S - S , we have:
2S = n•2n+1 + (n-1)•2n + (n-2)•2n-1 + ... + 3•24 + 2•23 + 1•22
-S=
- n•2n - (n-1)•2 n-1 - ... - 4•24 – 3•23 - 2•22 – 1•2 giving us
n
S = n•2
n+1
n
-2 -2
n-1
- ... - 2 - 2 - 2 = n•2
3
2
n+1
- 2i = n•2n+1 - (2n+1 - 2) = (n-1)• 2n+1 + 2
i=1
(Note: the sum of the powers of 2 is indexed starting with 1 not with 0 in this case.)
Finally, let Sn represent the sum of the first n terms of a series. Another method of finding a
closed form solution for the sum is to try to create an equation with Sn on both sides, but with
different coefficients so that we can solve for Sn. Let's evaluate the following which is a
generalization of the example we just did above:
n
Sn =  i•ai = a + 2a2+ 3a3 + ... + n•an. Then, adding the next term to the left hand side
i=1
n+1
Sn +
(n+1)•an+1
n
=  i•a = (i+1)•a
i
i=1
n
i=0
n
i+1
= i•a
i=0
n
i+1
+  ai+1
i=0
n
= a i•a + a  ai
i
i=0
i=0
= a•Sn + a[ (an+1 - 1)/(a - 1)] by formula 5),
so Sn = [a - (n+1)•an+1 + n•an+2] / (a – 1)2
Practice problems:
1. 2 + 4 + 6 + ... + 2n
2. 2 + 6 + 18 + ... + 2·3n
3. 2 + 6 + 10 + ... + (4n - 2)
4. 4 + 10 + 16 + ... + (6n - 2)
5. 1 + 5 + 9 + ... + 4n - 3
6. 1 + 3 + 6 + ... + [n(n + 1)]/2
7. 1•2 + 2•3 + 3•4 + ... + n(n + 1)
8. 1•3 + 2•4 + 3•5 + ... + n(n + 2)
2
2
2
2
9. 2 + 4 +6 + ... + (2n)
10. 12 + 32 + ... + (2n – 1)2
11. 1/(1•3) + 1/(3•5) + 1/(5•7) + ... + 1/[(2n - 1)(2n + 1)]
12. 1/(1•4) + 1/(4•7) + 1/(7•10) + ... + 1/[(3n - 2)(3n + 1)]
13. 2/(2•3) + 2/(3•4) + 2/(4•5) + ... + 2/[(n + 1)(n + 2)]
14. 1•2•3 + 2•3•4 + 3•4•5 + ... + n(n + 1)(n + 2)