Sequences And Series

Sequences and SeriesCBSE, Class XI

A Story About Gauss

When he was a child, to keep the class occupied, his teacher asked to sum all integers from 1 thru 100

Gauss solved it immediately. How?

Summing from 1 to 100Normal Sum = 1 + 2 + 3 + 4 + .... + 100

Reverse Sum = 100 + 99 + 98 + 97 + .... + 1

Normal + Reverse = 101 + 101 + 101 + 101 + .... + 101

• nth term of normal + nth term of reverse first term of normal + first term of reverse = 101

• we have 100 terms in each series• So Normal + Reverse = 101*100• Normal Sum = Reverse Sum (both have the same numbers) • Normal + Reverse = 2*Normal Sum = 101*100• Or Normal Sum = 101*100/2 = 5050

Key to summing Arithmetic Progressions

Lessons from Gauss' Story

Mechanical solving is not very useful

Insight is key to solving mathematical problems

Deutsche Mark is no longer in circulation. Euros replace Marks.

Gauss ideas still havent lost their currency. (His contributions still get high marks!)

While Money has currency, a Good Idea has permanence

Sequence•A sequence is an ORDERED list of objects

• Different from sets (in sets order does not matter)• The set A = {1,2,3,4,5} is equal to B = {5,4,3,2,1}

because each element of A Є B and each of B Є A

• The sequences S = 1,2,3,4,5 and T = 5,4,3,2,1 are different because ordering matters

Ordering => we can refer to nth element of a Sequence• We can say 5th element of a sequence of integers• Element can occur multiple times unlike for a set

Examples of SequencesA finite sequence has finite number of terms: e.g. 5,4,3,2,1

If a sequence is not finite it is called an infinite sequence • -1,1,-1,1,-1,1,-1,... first element is -1, second is 1, and nth

element is -1 to the power n

• 3,6,9,12,15,18... The nth element is 3*n

• 2,4,8,16,32,... The nth element is 2 to the power n

Example: Fibonacci SequenceKnown in India around 200 BC by Pingala who •discovered them in study of rhyme patterns in poetry •Also first wrote about binary numbers

Studied by Leonardo of Pisa, known as Fibonacci•All pairs born in year (n-1) will give birth in year n•All pairs born in year (n-2) will give birth in year n•All pairs born prior to n-1 will have died

births(n) = births(n-1) + births(n-2); the Fibonacci Sequence1,1,1+1=2,1+2=3,2+3=5,3+5=8,13,21,34,55,89,...

Fibonacci Sequence in Nature

Fibonacci numbers occur in nature in many places. Petals of flowers are an example

• http://library.thinkquest.org/27890/applications5.html• While individual examples may disagree, the most

common number of petals are usually close to Fibonacci numbers

Fibonacci Sequence: Ratio of TermsSequence = 1,1,2,3,5,8,13,21,34,55Ratio(k) = Term(k+1)/Term(k)r(1) = 1 / 1 = 1r(2) = 2 / 1 = 2r(3) = 3 / 2 = 1.5r(4) = 5 / 3 = 1.67r(5) = 8 / 5 = 1.6r(6) = 13 / 8 = 1.625r(7) = 21/13 = 1.615r(8) = 34/21 = 1.619r(9) = 55/34 = 1.617

Ratio of Successive Fibonacci No's

0

0.5

1

1.5

2

2.5

1 2 3 4 5 6 7 8 9 101112131415161718192021

Fibonacci Sequence Ratios: Converging Value

Lets say ratios converge to unknown value φ

So F(n+1)/F(n) = φ = F(n)/F(n-1) 1/φ = [F(n-1)/F(n)]

F(n+1) = F(n)+F(n-1) F(n+1)/F(n) = 1 + [F(n-1)/F(n)]

φ = 1 + 1/ φ or φ * φ = φ + 1 or ) φ = (1 + sqrt(5))/2

φ is irrational and often called Golden RatioAncient Greek fascination. Parthenon length/ht ~= φφ = 1.6180339887...

Sequences need not have well defined pattern

•For e.g. the sequence of primes 2,3,5,7,...

• There is no known formula for the nth prime number

• So this sequence has only a verbal description

Arithmetic Progressions

A sequence where, other than first term

every term = previous term + common difference

Notation:a = first term L = last term d = common difference S(n) = sum to n terms (Note: Slightly different notation from NCERT text)

General Term of Arithmetic Progression

1st Term = a = a + (1-1)*d

2nd Term = a + d = a + (2-1)*d

3rd Term = a + 2d = a + (3-1)*d

kth Term = a + (k-1)*d

If the series has n terms, then Last term is nth term and so

last term = L = a + (n-1)*d

Example:

First + Last Term = 80 for an AP with large no. of termsWhat is (a) Sum of 2nd term + last but 1 term (b) Sum of 3rd term + last but 2 term

2nd Term = First Term + common diff.Last but 1 Term = Last Term – common diff.2nd + Last but 1 Term = First Term + Last Term

3rd Term = 2nd Term + common diff.Last but 2 Term = Last but 1 Term – common diff.3rd + Last but 2 Term = 2nd Term + Last but 1 Term

= First Term + Last Term

So answer must be 80 for both questions.

Summing Arithmetic ProgressionsAP has the property

first term + last term =2nd term + last but 1 term =3rd term + 2nd before last term = a + (a+(n-1)*d) = (a+L)

1.normal sum = first + second + ... + last2.reverse sum = last + last but 1     + ... + first3.normal + reverse = (first + last) + (second + last but one) + ... + (last + first)4.normal + reverse = (first + last term)*(number of terms)5.normal sum*2 = (first + last term)*(number of terms)6.normal sum = (first + last term)*(number of terms)/2

Example: A.N. Kolmogorov at Age 5

Worked out sum of first n odd numbers for school journalfirst odd number = 1 = first term2nd odd number = 3 = 1 + (2-1)*2nth odd number = 1 + (n-1)*2 = last term

1,3,5,7,...till 1+(n-1)*2 is an AP with common difference 2 Summing this series yields an interesting answer Sum = [first term + last term]*(number of terms)/2

[1 + {1+(n-1)*2}]*n/2 = [1+1+2n-2]*n/2 = n*n

1+3 = 2*2; 1+3+5=3*3; 1+3+5+7=16=4*4;1+3+5+7+9=25=5*5

Example: Average of an APThe first term of AP is 11 and last is 56. We are not given( a ) common difference ( b ) number of termsCan we find the average? YesSum of AP = [Average of AP ]*{number of terms}Sum of AP       = [first + last term]*{number of terms}/2Average of AP = [first + last term]/2 (True For ANY AP)

So any AP starting with 11 and ending with 56 has average = (11+56)/2 = 33.5. This includes

AP1 = 11,16,21,26,31,36,41,46,51,56AP2 = 11,56AP3 = 11,12,13,....56

Example: 3 Consecutive terms of AP

Lets call the terms left, middle and right. Call the common difference d

Since they are in AP we know

1. middle = left + d or left = middle – d2. right = middle + d 3. left + right = 2*middle4.

left + middle + right = 3*middle

Example: Diff. of 2 terms of an APkth term = a + (k-1)*djth term = a + (j -1)*dkth term - jth term = (k –j)*d

depends only on d and (k-j) it does NOT depend on

• k and j but only on the difference k-j•first term (a)

15th term - 10th term = (15 -10)*d = 5d 10th term - 5th term = (10 - 5)*d = 5d 7th term - 2nd term = ( 7 - 2)*d = 5d

If 18th term - 10th term is 45, what is 20th term - 4th term? 90. (20-4)d = 16d = 2*(18-10)d=2*45

Arithmetic Mean (AM) of 2 NumbersGiven two numbers, the AM = 1/2 the sum

Example: Given 11,17 Arithmetic Mean = (11+17)/2 = 14.

A.M. has some interesting properties1. If the two numbers are equal, they are also equal to AM2. In case numbers are different

• AM is greater than smaller of the two numbers• AM is lesser than larger of the two numbers• AM - smaller number = larger number - AM = constant• Smaller number, AM, larger number are in AP with

common difference = constant in above equation

Create AP between 2 termsGiven first and last term, insert n terms to form APAP = first term, first term + d, first term + 2d, .... last term

Figure out the common difference d1. We know [last term - first term]2. n terms between first and last means last term is n+2th 3. In previous e.g., {n+2th term - first term} = (n+2-1)*d

4. d = [last term - first term]/(n+1)

Insert 9 terms between 2 and 122 so that they are in AP

• d = [122 - 2]/(9+1) = 120/10 = 12• AP is 2,14,26,38,50,62,74,86,98,110,122

Adding a constant to each term of APCall kth term of old AP, with common difference d, old(k)

old(k) = old(k-1) + d or old(k)-old(k-1) = d

Call kth term of new series as new(k) Given that new(k) = old(k) + constant

But is new series an AP? Yes, because1. new(k) = old(k) + constant2. new(k-1) = old(k-1) + constant3. new(k)-new(k-1) = old(k)-old(k-1) = d4. new(k) = new(k-1)+d5. new is AP with same common difference d as old AP

Multiplying a const. to each term of APCall kth term of old AP, with common difference d, old(k)

old(k) = old(k-1) + d or old(k)-old(k-1) = d

Call kth term of new series as new(k) Given that new(k) = old(k) * constant

But is new series an AP? Yes, because1. new(k) = old(k) * constant2. new(k-1) = old(k-1) * constant3. new(k)-new(k-1) = [old(k)-old(k-1)]*constant = d*constant

4. new(k) = new(k-1)+{d * constant}5. new is AP with common difference=d*constant

Geometric Progressions (GP)AP: any term = previous term + common difference (d)GP: any term = previous term * common ratio (r)

Above defines AP and GP for all terms except the first

AP: Kth term = (k-1)th term + dGP: Kth term = (k-1)th term * r

Examples of GP:1,2,4,8,16,32,64 Common ratio (r) = 21,1/2,1/4,1/8,1/16/,1/32 Common ratio (r) = 1/2

Nth term of GP

A bank pays annual interest of i% on the entire balance Deposit principal P at beginning of year and no withdrawals

What is his balance at beginning of each year?•1st year = P •2nd year = P*(1+i) •3rd year = P*(1+i)*(1+i) •4th year = P*(1+i)*(1+i)*(1+i) = P*[(1+i) ]3 •nth year = P*[(1+i) ]n-1

For a GP beginning with a and common ratio rnth term = a * [ r ]n-1

Note: nth term = (n-1)th term * r

Sum to n terms of GP

If r is 1, every term = first term, sum = n * first term = n*a S(n) = (a) + (a*r) + (a*r*r) + ... (a*rn-1)

S(n)*r = (a*r) + (a*r*r) + ... (a*rn-1) + (a*rn)

1. every term in RHS of S(n) is in S(n)*r except (a)2. S(n)*r has additional term (a*rn)3. S(n)*r = S(n)-a + (a*rn)

4. S(n)*[r-1] = a* [ rn - 1 ] = [last term* r– first]

5. S(n) = {a/(r-1)} * [ rn - 1   ]

Doesn’t depend on n Although Last term a*rn-1

Sum depends on rn

= [1/(r-1)]*[last term * r – first]

Example: Summing a GP and Nth Term1+2+4+8+... to n terms. This GP has interesting properties

S(n) = 1/(r-1)*[(last term * r) – first term] r = common ratio = 2So (1/(r-1)) = 1first term = 1 

S(n) = {(last term * 2)-1 }

1+2 = ( 2*2) – 1 = 31+2+4 = ( 4*2) – 1 = 71+2+4+8 = ( 8*2) – 1 = 151+2+4+8+16 = ( 16*2) - 1 = 311+2+…+512 = (512*2) – 1 = 1023

Example: 3 Consecutive terms of GPCall these the left term, middle term, and right term

middle term = left term * r left term = middle term / rright term = middle term * r

left term * right term = middle term * middle term

Product of these three terms isP = (left term)*(middle term)*(right term)P = cube of the middle term

Also, (middle term/left term) = (right term/middle term) = r

Example: Ratio of any 2 terms of GPkth term = a * rk

jth term = a * rj kth term/jth term = r (k-j)

Observe that the ratio depends on • r• k – j

Ratio does not depend on • the first term a• individual values of k, j but depends only on k - j

So for example, for any GP, ratio of18th Term/7th term = r11

27th Term/16th term = r11

Create GP between 2 termsGiven first and last term, insert n terms to form GPGP = first term, first term*r, first term*r*r, .... last term

Figure out the common difference ratio r1. We know [last term - first term]2. n terms between first and last means last term is n+2th 3. In previous e.g., {n+2th term/first term} = r (n+2-1)

4. r = [last term/first term] (1/[n+1])

Insert 3 terms between 3 and 243 so that they are in GP

• r = [243/3] (1/[3+1]) = 81 (1/4) = 3• GP is 3,9,27,81,243

Geometric Mean (GM) of 2 NumbersGM is square root of the product of the two given numbers

Example: Given 4,9 GM = square root of 36 = 6

G.M. has some interesting properties1. If the two numbers are equal, they are also equal to GM2. In case numbers are different, call them Big, Small

• GM = sqrt(Big*Small)• GM/Small = sqrt(Big*Small)/Small = sqrt(Big/Small) > 1

• GM is greater than Small

• Big/GM= Big/sqrt(Big*Small)/ = sqrt(Big/Small) > 1• GM is lesser than Big

• (GM/Small number) = (Big number/GM)• Small number, GM, Big number are in GP

Relationship Between GM and AM

Two numbers s and L, both >= 0, AM = (s+L)/2, GM = sqrt(s*L)

If both numbers are equal AM = GM = s = LIf not lets call the larger of the two numbers L

If s is really small, say s = 0, GM = 0 but AP = L/2 > 0So we guess that AM >= GM

AM-GM = (s+L)/2 - sqrt(s*L) = [s + L - 2*sqrt(s*L))] /2 = [sqrt(L) - sqrt(s)] /2

The RHS of last equation is >=0 so AM-GM >=0