Series Definition & Meaning

QUESTION 1
(a)
(i) Series C
(ii) Series B
(iii) Series H
(b)
(i) Series C the first term is 6
(ii) Series C: Common difference (d) = 7 – 6 = 8 – 7 = 9 – 8 = 10 – 9 = 1
(iii) The nth term, an is given by
an=a1 + ( n−1 ) d
Where a1 is the first term.
For series C, the 10th term is
a10=6+9 d =15
(iv) The sum of the first 10 terms is given by
¿ n
2 [ 2 a+ ( n−1 ) d ]
For series C, the sum is
¿ 10
2 [ 2(6)+ ( 10−1 ) ]=105

(c)
(i) Series B the first term is 4
(ii) Series B: Common ratio (r) is
¿ 2
4 =1
2 = 0.5
1 = 0.25
0.5 =0.5
(iii) The nth term, an is given by
an=a1 rn−1
Where a1 is the first term.
For series B, the 8th term is
a8=4 ( 0.5 )7= 1
32 =0.03125
(iv) The sum of the first 8 terms is given by
Sn= a1 (rn−1)
r−1 , for r >1
Or
Sn= a1 (1−rn )
1−r , for r <1
For series B, the sum is
¿ 4 (1− ( 0.5 ) 8)
1−0.5 =7.96875

(v) By geometric series test:
Series B converges since the absolute value of the common ratio is less than unity i.e.
r =1
2 which is ¿ 1
(d)
Series H is not Arithmetic Progression (AP) since there exist no common difference (d)
between the consecutive terms i.e.
Series H: (4 – 1) ≠ (9 – 4) ≠ (16 – 9) ≠ (25 -16)
Series H is not Geometric Progression (AP) since there exist no common ratio (r)
between the consecutive terms i.e.
For series H: 4/1 ≠ 9/4 ≠ 16/9 ≠ 25/14
QUESTION 2
The tunneling machine follows an Arithmetic Progression (AP) in drilling. This progression is
defined by the model:
Cost of drilling=X + ( 500−1 ) Y
For X = 1200 and Y =100, the total cost of drilling is
¿ 1200+100 ( 500−1 )
¿£ 51100

QUESTION 3
a)
From the general relation of Geometric Progression T n=a r n−1
First term: T 1=a=X=14
Fifth term: T 5=a r 4=224
From the fifth term expression r =( 224
a ) 1
4
But a=14 from the first term expression
This implies that r =( 224
14 )1
4 =2
∴ Commonratio=2
b)
Term Expression Speed
First T 1=a r0 =14 ( 2 )0 14 rpm
Second T 2=a r1=14 ( 2 )1 28 rpm
Third T 3=a r2 =14 ( 2 )2 56 rpm
Fourth T 4=a r3=14 ( 2 )3 112 rpm
Fifth T 5=a r 4=14 ( 2 ) 4 224 rpm

QUESTION 4
a)
The value after Y years ¿ P (1− X
100 )Y
Where P = Initial Value
& X = Depreciation rate
For P=£ 100000 X =1.4∧Y =2
The value after 2 years ¿ £ 100000 ( 1− 1.4
100 )
2
=£ 97219.60
b)
From the relation Final value=Initial value ( 1− X
100 )
Period
Period=
log ( Final value
Initial value )
log (1− X
100 )
→ Period=
log ( 50000
100000 )
log (1− 1.4
100 ) =49.16 years
∴ Period=49.16 years

QUESTION 5
The Pascal triangle for a function raised to power three is as shown below.
From the triangle, the co-efficients are derived as:
i) Only number ‘1’ exist on the outside
ii) Each of the remaining numbers in the middle is obtained from addition of two
numbers just above. Below is a representation
For any natural number n and any binomial a + b,
(a + b)n = k0anb0 + k1an-1b1 + k2an-2b2 + …+ kn-1a1bn-1 + kna0bn where the constant numbers kn, kn-1 ,
…, k2, k1, k0 are obtained from the (n+1)st row of the Pascal triangle.
Therefore, to expand (1 + 2x)3; a is set to 1, b is set to 2x and n is set to 3. 4th row of the Pascal
triangle is used. The co-efficients derived from the 4th row are: 1 3 3 1
Thus,
( 1+2 x ) 3 =1 ( 1 ) 3 ( 2 x ) 0 +3 ( 1 ) 2 (2 x)+3 ( 1 ) 1 ( 2 x )2 +1 ( 1 ) 0 ( 2 x ) 3
¿ 1+6 x+12 x3 +8 x3

QUESTION 6
Generally for any natural number n, the binomial expansion of (a + b)n is given by
( a+ b )n=∑
k=0
n
(n
k )an−k bk
¿ (n
0 )an b0 +(n
1)an−1 b1 + (n
2 )an−2 b2 +…+ ( n
n−1 )a1 bn−1 +(n
n)a0 bn
Comparing ( a+ b ) n with ( 1+2 x ) 3 ; a is set to 1, b is set to 2x and n is set to 3. So
( 1+2 x ) 3 =∑
k=0
3
( 3
k ) (1)3−k (2 x )k
¿ ( 3
0 ) ( 13 ) + ( 3
1 ) ( 12 ) ( 2 x ) +( 3
2 ) ( 11 ) ( 2 x ) 2 +( 3
3 ) ( 2 x ) 3
¿ 3 !
0 !3! + 3 !
1 !2! ( 2 x ) + 3 !
2! 1 ! ( 4 x2 ) + 3 !
3 !0 ! ( 8 x3 )
¿ 1+6 x+ 12 x2 +8 x3
QUESTION 7
The Binomial expansion of ( 1+ x )−n is generally given by the expression below
( x +a )−n=∑
k=0
n
(−n
k )a−n −k xk
¿ ∑
k=0
∞
( −1 ) k
( n+k −1
k ) xk a−n−k
For |x|<1,
( 1+ x )−n=1−nx+ n ( n+1 )
2 ! x2 −n ( n+1 ) ( n+2 )
3 ! x3 + …+ ( −1 ) r n ( n+ 1 ) ( n+2 ) … ( n+ r−1 )
r ! xr +…
and so,

( 1+ x )−3=1−3 x + 3 ( 3+1 )
2! x2− 3 ( 3+1 ) ( 3+2 )
3 ! x3 +…¿ 1−3 x +6 x2−10 x3 +…
This expansion holds for values of x such that x is a real number and |x|<1
QUESTION 8
Given a function f(x) = ex
The Maclaurin series expansion of a general function f(x) is given by
f ( x )=f ( 0 ) +x f ' ( 0 ) + x2
2 ! f ' ' ( 0 )+ x3
3 ! f ' ' ' ( 0 ) + …
For f ( x ) =ex,
f ( 0 )=e0=1
f ' ( x )=e x → f ' ( 0 )=e0=1
f ' ' ( x ) =ex → f '' ( 0 ) =e0 =1
f ' ' ' ( x )=ex → f ' '' ( 0 )=e0 =1
This implies that ex=1+x ( 1 ) + x2
2 ! ( 1 )+ x3
3 ! ( 1 )+ …
∴ ex=1+x + x2
2! + x3
3 !+ …

QUESTION 9
From the above Maclaurin series expansion of ex
For x = 2,
e2 ≈ 1+2+ 22
2! + 23
3 !
≈ 19
3 ∨6.333
Actual value of ex=e2 =7.3891. Therefore, expansion by Maclaurin series closely approximates
the exact solution.
For question 7, when x = 2,
¿ 1−3 x +6 x2−10 x3 +…=−61
While the actual value is
( 1+x )−3= ( 1+2 ) −3= 1
27
A wide deviation exists in this case since the value of x chosen is outside the valid range of
expansion i.e. the absolute value of x should be less than unity for the expansion to be valid.

QUESTION 10
Given that T original =2 π √ L
g
An increase in the length of the pendulum by X% results into a new length equivalent to
(1+X/100)*L
For X% = 1%, the new length is Lnew=1.01L
→ Tnew=2 π √ 1.01 L
g
Considering that Error = Tnew - Told
error =2 π √ 1.01 √ L
g −2 π √ L
g
¿ 2 π √ L
g [ √ 1.01−1 ]
Expressing the error as a percentage of the original period
%error = error
T original
x 100 %
%error =
2 π √ L
g [ √1.01−1 ]
2 π √ L
g
x 100 %
¿ [ √1.01−1 ]∗100 %
¿ 0.4988 %≅ 0.5 %
Therefore %error ≅ 0.5 %