Series and Sequences

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Sequences (Nth term)

Sequences are sets of ranges which follow the pattern

U1, U2, U3, U4 .... Un

3 , 7, 11, 15 ... Un

+4 +4 +4 is the common difference

So we can say 4n is the basis of the sequence

4, 8. 12. 16 ... Un

3 , 7, 11, 15 ... Un

-1 -1 -1 common difference of the 2 sequences

So we can say the Un = 4n-1

Sequences (Recurrent relationships)

U1 = 6 U1, U2, U3, U4

U2 = 13 6, 13,27,55

U3 = 27 the common difference is the previous term x2 and +1

U4 = 55 So when n ≥1 and we known the first term U1 = 6 we can derive the formula -----------------> Un+1 = 2(Un)+ 1

So to work out U5 we would have to plug U4 into our derived formula

U(4+1) = 2(U4) + 1

U5 = 2(55) +1

U5 = 111

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Arithmetic sequences

Arithmetic sequences always have commas

6, 8,10,12

+2 +2 +2 (common difference which is increasing) [The common difference is always the same consecutive number 11,8,5,2 If the CD is not consecutive it cannot be an arithmetic sequence]

-3 -3 -3 (common difference which is decreasing)

a represents the first term, and d represents the common difference

In general U1= a U2= a + d U3= a+2d [a is always the first term of the sequence]

So we can conclude the formula for finding the nth term would be Un = a + (n-1)d

Arithmetic series

There are 2 formulas:

Sn = (n/2)[2a + (n-1)d]

Sn = (n/2)[2a + (a+l)] (This can only be used if we are given, or know the last term. Can use Un = a + (n-1)d to find terms)

Arithmetic series always have plus'

E.G: Find the sum of the first 21 terms in the sequence below

5,9, 13, 17

+4 +4 +4 (common difference which is increasing) [We know the common difference is 4 as it goes up consecutively]

S21 = (21/2)[2(5)+ (20)(4)] = 21[(10 + 80)/2] [by 4 each term, and the a is 5] as it is the first term and n is 21

S21 = (21)(45) = 945 [because we are trying to find the sum of the first 21]

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Geometric sequence

In a geometric sequence a is always the first term of the sequence, and the common ratio, r.

So we can conclude that the nth term formula is, Un = ar^(n-1)

E.G: Find the 6th term of the sequence below below.

4 , 12 , 36 , 108 [We know that a is 4, this is because it is the first term in the sequence, we also

x3, x3, x3, x3 know r is x3 because it is multiplying by 3 each time]

Un = ar^(n-1) so Un = (4)(3)^(6-1)

U6 = (4)[(3)^5] = 4[243] = 972

Geometric series

In a geometric sequence a is always the first term of the sequence, and the common ratio, r.

So we can conclude that the nth term formula is, Sn = a[(r^n) - 1] or Sn = a[(1 -(r^n)]

(r - 1) (r - 1)

E.G: Find the sum of the first 10 terms from the sequence below.

4 , 12 , 36 , 108 [We know that a is 4, this is because it is the first term in the sequence, we also

x3, x3, x3, x3 know r is x3 because it is multiplying by 3 each time]

Sn= a[(r^n) - 1] so S10= (4)[(3)^(10)] - 1]

(r - 1) (3 - 1)

S10 = 118096

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Sum to infinity

Sn = a[(1 -(r^n)] (r - 1)

If -1<r<1 and n is a large number e.g 1000 then r will converge and tend towards 0 this means that we can ignore the bracket because (1-0) = 1 and 1a = a. Based on this we can derive a new formula wihich would be

S∞ = a/(r-1) [THIS ONLY WORKS IF |r|<1]

Sigma notation

Sigma sign is used to show the sum of a series (all the desired number from the series added together)

7 [Because we have r=3 we would sub 3 in to find

Σ (2r+3) = 9 + 11 + 13 + 15 + 17 = 65 our first term being 9. We would sub values

r = 3 3 to 7. Then add them all together.

Two important sigma formulas:

n n n

Σ af(r) = aΣ f(r) Σ a = a + a + a + a + ..... + a

r = x r = x r = 1 = n(a)

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