Sequences and Series: Finite Series and Binomial Theorem, Summaries of Computer Science

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Sequence & Finite Series

B.E Owusu

KNUST

February 6, 2023

Series Notation

A sequence is an ordered set of numbers that most often follows some rule (or pattern) to determine the next term in the order. For example, x, x^2 , x^3 , x^4 , ... is a sequence of numbers, where each successive term is multiplied by x. A series is a summation of the terms of a sequence. The greek letter sigma is used to represent the summation of terms of a sequence of numbers. Series are typically written in the following form:

∑^ n

i = 1

ai = a 1 + a 2 + a 3 .... + a 4

Examples (a) ∑^5

i = 1

i = 1 + 2 + 3 + 4 + 5 = 15

(b)

∑^5

k = 3

2 k^ = 23 + 24 + 25 + 26 = 8 + 16 + 32 + 64 = 120

(c)

∑^5

n = 1

2 k^ = 11 + 22 + 33 + 44 = 1 + 4 + 27 + 256 = 288

Types of Sequences & Series

There are two main types of sequences. An arithmetic sequence is one in which successive terms differ by the same amount. For example, 3, 6, 9, 12, ... is an example of an arithmetic sequence, where each term is obtained by adding 3 to the previous term. A geometric sequence is one in which the quotient of any two successive terms is a constant. For example, 3, 9, 27, 81, ... is an example of a geometric sequence, where each term is obtained by multiplying the previous term by 3. Similarly, there are also arithmetic series and geometric series, which are simply summations of arithmetic and geometric sequences, respectively.

Theorems of Finite Series

The following theorems give formulas to calculate series with common general terms. These formulas, along with the properties listed above, make it possible to solve any series with a polynomial general term, as long as each individual term has a degree of 3 or less.

Euler’s Formula ∑^ n

i = 1

1 = n

∑^ n

i = 1

c = nc

∑^ n

i = 1

i =

n ( n + 1 ) 2

∑^ n

i = 1

i^2 =

n ( n + 1 )( 2 n + 1 ) 6

∑^ n

i = 1

i^3 =

( n ( n + 1 ) 2

) 2

Computing Finite Series

The following theorem is used for calculating arithmetic series. Suppose we have the following arithmetic series, { a + ( a + d ) + ( a + 2 d ) + ... + ( a + ( n − 1 ) d }. Then,

n ∑ − 1

i = 1

a + kd =

n 2

( 2 a + ( n − 1 ) d )

The following theorem is used for calculating geometric series. Suppose we have the following geometric series, { a + ar + ar^2 + ... + ar ( n −^1 )}. Then,

n ∑ − 1

i = 1

ar k^ = a

( r n^ − 1 r − 1

)

Example 2 Find the sum of the geometric series: 2 + 8 + 32 + 128 + ... + 8192

Solution

We know that

= 4 and that

= 4 , so r = 4 for this geometric

series. The initial value represents our a value, so a = 2. Before we can write the series in summation notation, we must determine the upper limit of the summation.

8192 = ar n −^1 = ( 2 )( 4 n −^1 ) 4096 = 4 n −^1 46 = 4 n −^1 6 = n − 1 n = 7

So, we can rewrite the series as

∑ 6 k = 0 2 (^4

k (^) ). From the formula for

the sum of a geometric series,

∑^6

k = 0

2 ( 4 k^ ) = 2

( ( 4 )^7 − 1 4 − 1

)

Therefore, the sum of the series, 2 + 8 + 32 + 128 + ... + 8192 is 10922

Example 1 Use the Binomial Theorem to expand ( 2 x − 4 )^4

Solution

( 2 x − 3 )^4 =

∑^4

i = 0

( 4 i

) ( 2 x )^4 − i^ (− 3 ) i

( 4 0

) ( 2 x )^4 +

( 4 1

) ( 2 x )^3 (− 3 ) +

( 4 2

) ( 2 x )^2 (− 3 )^2 + ...

= ( 2 x )^4 + 4 ( 2 x )^3 (− 3 ) +

( 2 x )^2 (− 3 )^2 + 4 ( 2 x ) ...

= 16 x^4 − 96 x^3 + 216 x^2 − 216 x + 81

Binomial Series

If k is any number and | x | < 1 then,

( 1 + x ) k^ =

∑^ ∞

n = 0

( k n

) xn

= 1 + kx +

k ( k − 1 ) 2!

x^2 +

k ( k − 1 ) ( k − 2 ) 3!

x^3 + · · ·

where, ( k n

) =

k ( k − 1 ) ( k − 2 ) · · · ( k − n + 1 ) n****!

n = 1 , 2 , 3 ,... ( k 0

) = 1

Range of Validity of a Binomial Expansion

As stated above, the second formula for binomial expansion valid for | x | < 1. This is because, unlike for positive integer n , these expansions have an infinite number of terms (as indicted by the

... in the formula). Subsequently, we require the series to converge as the powers of x become large. For this to happen, we must have | x | < 1 Also notice that in this second formula there is a very specific format inside the brackets – it must be 1 plus something. Therefore, if there is something other than 1 inside these brackets, the coefficient must be factored out. Do this by first writing

( a + bx ) n^ =

( a

( 1 +

bx a

)) n = an

( 1 +

bx a

) n

Then find the expansion of

( 1 +

bx a

) n

using the formula. Do this by replacing all x with

bx a

This

inevitably changes the range of validity. It follows that this expansion will be valid for

∣∣ ∣ bxa

∣∣ ∣ < 1 or | x | < ab

The first four terms in the binomial series is then,

√ 9 − x = 3

( 1 +

( −

x 9

)) (^12)

∑^ ∞

n = 0

( (^1) 2 n

)( −

x 9

) n

  1 +

( 1 2

) ( −

x 9

) +

1 2

( −^12

)

( −

x 9

) 2 +

1 2

( −^12

) ( −^32

)

( −

x 6

x^2 216

x^3 3888