SEQUENCES AND SERIES Q 11B
For a certain infinite series, the ratio lim
n 
U n 1
Un

x
. The series will definitely converge under which of
2
the following conditions?
X
A:
x2
X
B:
x2
X
C:
x 2
Y
D:
x 2
X
E:
x2
A:
Incorrect.
B:
Incorrect.
The required condition is x  2 so that the limiting value of the ratio is <1
The required condition is x  2 so that the limiting value of the ratio is <1.
This value makes the ratio 1 and then this test fails.
The required condition is x  2 so that the limiting value of the ratio is <1.
C:
Incorrect.
D:
Correct.
This is the required condition so that lim
E:
Incorrect.
No, the terms are getting bigger as n increases.
When x  2 , the ratio is 1 and then this test fails
n
U n1
Un
1
SEQUENCES AND SERIES Q 12
Which expression is correct for the general kth term of the sequence: -1, cos1, - cos2, cos3, ..........
Consider carefully at what value k starts, 0 or 1.
Y
A:
 1k cos(k  1)
X
B:
 1k cos(k )
X
C:
 1k 1 cos(k  1)
X
D:
1k 1 cos(k  1)
X
E:
1k cos(k )
A:
Correct.
This is correct for k  1 but if k started at zero, the first term
would be +cos(-1) which equals cos(1).
B:
Incorrect.
This does not work even if k started at 0. See below for more explanation.
C:
Incorrect.
Here k would need to start at 1 to give cos(0)=1 but (-1)0=+1. Therefore the first
term would have the wrong sign. See below for more explanation.
D:
Incorrect.
The (+1)k-1 will not produce the required sign changing.
See below for more explanation.
E:
Incorrect.
The (+1)k will not produce the required sign changing.
See below for more explanation.
The term (-1) is included in the general term to produce the changing sign as k increases:
this eliminates answers D and E. For k =0, A, B and C give us +cos(-1) and +cos(0) and -cos(-1)
respectively as their first terms. However starting at k =1, answer A gives the required sequence
commencing at -1, whereas the first terms of Band C are -cos(1) and = +cos(-1) respectively.
SEQUENCES AND SERIES Q 13
Which expression is correct for the sum to infinity of the series: -1+ cos1 - cos2 + cos3 ......

Y
A:
  1
k
cos(k  1)
k
cos(k  1)
k 1

X
B:
  1
k 0

Y
C:
  1
k 1
cos(k )
k 0

X
D:
  1
k
cos( k )
k 1

X
E:
  1
k
cos( k )
k 0
A:
Correct.
This works since k starts at 1, but another answer also works! See below.
B:
Incorrect.
Starting at k =0 gives a first term of cos(-1) not just -1. See explanation below.
C:
Correct.
This is correct as well as answer A. k =0 gives the first term -1,
k =1 gives +cos1 etc. See explanation below for more details.
D:
Incorrect.
The cos(k) term gives cos1 for the first term.
See explanation below for more details.
E:
Incorrect.
The cos(k) term gives cos1 for the first term.
See explanation below for more details.
Starting the summation at k =0, gives a (+1) term (i.e. the wrong sign), for the first term of answers B and
E, but is OK for answer C. Remember (-1)0=+1 and cos (-1)=cos (1). Starting at k =1, answers D
has -cos(1) as its first term, but answer A is correct as is answer C.
SEQUENCES AND SERIES Q 14
You may use a calculator for this problem
Given sin

4

1
2
 0.7071 and the correct value for sin 46.80  0.72897 , the calculated value of

sin 46.80 found using only the first three terms in a Taylors expansion, starting at
X
A:
0.6852
X
B:
0.7071
X
C:
0.7293
Y
D:
0.7290
X
E:
0.7297
A:
Incorrect.
The value of sin 46.50 cannot be less that the value of sin
4

4
, is;
. You made a mistakes
in the signs of the second and third terms when you differentiated..
The correct work is shown below:

i.e sin 450 . The correct work is shown below:
B:
Incorrect.
This is the value of sin
C:
Incorrect.
You have only calculated the first two terms. The correct work is shown below:
D:
Correct.
You have calculated all three terms with the correct signs.
E:
Incorrect.
You have calculated all three terms but got one sign incorrect when you
differentiated cosine. The correct work is shown below:
4
Remember Taylor's series f ( x  h)  f ( x)  hf '( x) 
Here; sin

 sin 450 , h=
4
2







f( 
) f( )
f '( ) 
f ''( )
4 100
4 100
4
20000
4
Inserting values:

h2
f ''( x) to three terms.
2!
 1.80
100





2

 sin( 
)  sin( ) 
cos( ) 
sin( )
4 100
4 100
4
20000
4
sin 46.80  0.7071(1  0.0314  0.0005)  0.7071(1.0309)  0.7290 (by calculator)
SEQUENCES AND SERIES Q 15
Insert 3 terms between 64 and 4 to form an geometric series. The sum of these three terms of this series
is:
X
A:
68
X
B:
124
Y
C:
-24
X
D:
44
X
E:
-20
A:
Incorrect.
This is the sum of the given first and last terms. You need to work out the 3
missing terms. The correct work is shown below:
B:
Incorrect.
This is the sum of all 5 terms. You have also missed the clue in answers C and E
(which also happens to be the wrong value by the way)
The correct work is shown below:
C:
Correct.
Well done. You spotted the common ratio could be negative.
The detailed work is shown below:
D:
Incorrect.
You spotted the common ratio could be negative but summed all 5 terms
The correct work is shown below:
E:
Incorrect.
You spotted the common ratio was negative but included the fifth term.
The correct work is shown below:
The nth term in a geometric series is ar n 1 where a is the first term and r is the common ratio. For the last
term r = 4
 64r 4  4
r 4 
1
1
giving r  
16
2
Try both values and the only match is answer C with
r negative. Now calculate the value of the second term. It is -32! The third is +16 and the fourth is -8.
Therefore for the middle three terms, starting at -32:
 1
9
32 1   32  
a(1  r )
 8 
 8   24
Sn 

1
3
1 r
1
2
2
n
SEQUENCES AND SERIES Q 16
In the binomial expansion (a  b)n the expression for the general rth term is:
Cr 1a n r 1br 1
Y
A:
n
X
B:
n
Cr a r bn
X
C:
n
Cr a n br
X
D:
n
Cnr a nr br
X
E:
n
Cr a n r b r
A:
Correct.
Notice in the expansion below, the third term the number 2 occurs, so the rth term
involves r-1 :
B:
Incorrect.
The b n occurs in the last term of the expansion and the coefficient n Cr is that for
the r+1 term The expansion is shown below:
C:
Incorrect.
The a n occurs in the first term of the expansion and the coefficient n Cr is that for
the r+1 term The expansion is shown below:
D:
Incorrect.
Although the coefficient is the same as n Cr (remember n Cr n Cnr ) it is the one for
the r+1 term. The expansion is shown below:
E:
Incorrect.
This is actually the r+1 term. The expansion is shown below:
(a  b)n  nC0 a n  nC1a n 1b  nC2 a n2b2  .........  nCr 1a n r 1br 1  nCr a n r br  ........  nCn1abn 1  nCnbn
The rth term is nCr 1a nr 1br 1 since r starts from zero
SEQUENCES AND SERIES Q 17
The sum to n terms of the binomial expansion (a  b)n is given by:
X
A:
Sn 
n
 (a  b)
n
r 1
X
B:
Sn 
n

n
Cr 1a n  r 1b r 1
r 1
X
C:
Sn 
n

n
Cr a n  r 1b r 1
r 1
X
D:
Sn 
n

n
Cr a n  r b r
n
Cr a n  r b r
r 1
Y
E:
Sn 
n

r 0
A:
Incorrect.
There are two things wrong with this answer. Firstly r does not appear in the
general term inside the sigma, and secondly there is no general term at all!
B:
Incorrect.
The power of a is wrong. For r =1 we get an+2 rather than the required an.
C:
Incorrect.
The coefficient n Cr is wrong. For r =1, i.e. the first term, we get n not 1.
It would be OK if n Cr was replaced by n Cr 1 .
D:
Incorrect.
This would work if r started at 0.
E:
Correct.
Notice that here r starts at zero.
The binomial expansion is:
(a  b)n  nC0 a n  nC1a n 1b  nC2 a n2b2  .........  nCr 1a n r 1br 1  nCr a n r br  ........  nCn1abn 1  nCnbn
The general term is: nCr a nr br with r starting at zero.
SEQUENCES AND SERIES Q 18
Which of the following series has the general term (1)k k where k is an integer?
Y
A:
-1+2-3+4-5+
X
B:
1+2+3+4+5+
X
C:
0+1-2+3-4+5-
X
D:
1+2+4+6+8+
X
E:
-1+2-4+6-8+
A:
Correct.
The minus 1 term changes sign as k increases and k starts at1.
B:
Incorrect.
The minus 1 term changes sign as k increases.
C:
Incorrect.
The minus 1 term changes sign as k increases, but produces +1 when k is even
and in this case k would start at 0.!
D:
Incorrect.
This would be true if k only took even values.
E:
Incorrect.
For this to be the answer k would have to only be even values and the power
outside the bracket would be k/2 or the k term would have to be doubled to 2k
SEQUENCES AND SERIES Q 19
Which value is correct for the sum to five terms of the expansion of (1+3)4
X
A:
1024
X
B:
512
Y
C:
256
X
D:
175
X
E:
81
A:
Incorrect.
You do not need to expand the binomial to work this out. It is 44 of course!
B:
Incorrect.
You do not need to expand the binomial to work this out. It is 44 of course!
C:
Correct.
You do not need to expand the binomial to work this out. It is 44 of course!
D:
Incorrect.
I think you have performed a binomial expansion and forgotten the last term.
You do not need to expand the binomial to work this out. It is 44!
E:
Incorrect.
You do not need to expand the binomial to work this out. It is 44 not 34!
SEQUENCES AND SERIES Q 20
Which is the correct expression for the Maclaurin series for f(x)?:
X
A:
f ( x)  hf '( x) 
h2
h3
f ''( x) 
f '''( x) 
2!
3!
Y
B:
f (0)  xf '(0) 
x2
x3
f ''(0) 
f '''(0) 
2!
3!
X
C:
1 x 
X
D:
1
x2 x4


2! 4!
X
E:
x
x3 x5


3! 5!
A:
Incorrect.
This is the Taylor expansion. Maclaurin is a special case of this.
B:
Correct.
If you compare this with the Taylor expansion you can see the similarity.
C:
Incorrect.
This is the power series for ex. It can be derived using the Maclaurin expansion.
Try to remember it since it is very important!
D:
Incorrect.
This is the specific case of the power series for cosx. . It can be derived using the
Maclaurin expansion.
E:
Incorrect.
This is the specific case of the power series for sinx. . It can be derived using the
Maclaurin expansion.
x 2 x3


2! 3!
SEQUENCES AND SERIES Q 21
Which is the correct expression for the Taylor series for f(x+h)?:
Y
A:
f ( x)  hf '( x) 
h2
h3
f ''( x) 
f '''( x) 
2!
3!
X
B:
f (0)  xf '(0) 
x2
x3
f ''(0) 
f '''(0) 
2!
3!
X
C:
1 x 
X
D:
1
x2 x4


2! 4!
X
E:
x
x3 x5


3! 5!
A:
Correct.
This is the Taylor expansion. (Maclaurin is a special case of this).
B:
Incorrect.
This is the Maclaurin expansion. If you compare this with the Taylor expansion
you can see the similarity.
C:
Incorrect.
This is the power series for ex. Try to remember it since it is very important!
D:
Incorrect.
This is the specific case of the power series for cosx.
E:
Incorrect.
This is the specific case of the power series for sinx.
x 2 x3


2! 3!
SEQUENCES AND SERIES Q 22
The sum of the first three numbers of an arithmetic series is 33 and their product is 792. What is the sum
of the first 4 numbers in this series?
X
A:
792
Y
B:
58
X
C:
42
X
D:
46
Y
E:
30
A:
Incorrect.
The only way that the sum of any numbers can be the same as their product is if the
numbers are all 0's or two 2's.
B:
Correct.
You have found the three values and the correct common difference. Did you
realise that it could have been -7 as well, so answer E is also correct? The full
working is below:
C:
Incorrect.
It looks like you have found the correct value for the middle number of the three
but got the wrong value for the common difference. That is the same mistake as for
answer D Weird!!! The correct working is below:
D:
Incorrect.
It looks like you have found the correct value for the middle number of the three
but got the wrong value for the common difference. That is the same mistake as for
answer C. Weird!!! The correct working is below:
E:
Correct.
You have found the three values and the correct common difference. Did you
realise that it could have been +7 as well, so answer B is also correct? The full
working is below:
Let x be the middle of the three numbers and d be the common difference. The sum is then:
(x-d)+x+(x+d)=33
therefore x=11. The product: (x-d)x(x+d)=792 which gives d2=49 i.e.d=+7 or -7
The possible first four terms are: 4+11+18+25=58 or 18+11+4-3=30
SEQUENCES AND SERIES Q 23
Insert five numbers between 6 and 66 to form an arithmetic series. The sum of the middle three terms is:
X
A:
36
Y
B:
108
X
C:
126
X
D:
180
X
E:
252
A:
Incorrect.
This is the value of the middle term but you need to find the common
difference as well. See below for the answer.
B:
Correct.
Yes you have found the common difference as 10 and added the three middle terms
of 26, 36 and 46
C:
Incorrect.
I believe you have calculated the wrong value for the common difference and made
it 12 and therefore only have six numbers in the series, rather than seven. Finally
you have chosen three near the middle to add together. I think we can call this
definitely WRONG.
D:
Incorrect.
I think you have found the correct common difference of 10 but added the five
middle terms instead of three by mistake. See below for the answer.
E:
Incorrect.
Looks like the common difference value is correct at 10, but you have added all
seven terms. See below for the answer.
The last term is given by l  a  (n  1)d Inserting numbers: 66  6  (6)d . Remember n=7 in total.
This solves to give d =10. Therefore the series is: 6+16+26+36+46+56+66 and the required sum=108.
SEQUENCES AND SERIES Q 24
Which formulae are correct for the sum of the first n natural numbers (i.e 1+2+3+4+5 ....)?
1 d n 

 1 d 
(i): 
(ii):
n(n  1)(2n  1)
6
(iii):
n
2a  (n  1)d 
2
n
2
(iv): (n  1)
(v):1+(n-1)
X
A:
(i) and (ii)
X
B:
(i) and (v)
Y
C:
(iii) and (iv)
X
D:
(iii) and (v)
X
E:
(ii) and (iv)
A:
Incorrect.
Formula (i) is for the sum of a geometric progression with the symbol d replacing
the normal r for the common ratio. For the given series d is actually 1 but this
formula becomes 0/0. Formula (ii) is the sum to n for squares of the first n natural
numbers, i.e 12+22+32+...
B:
Incorrect.
Formula (i) is for the sum of a geometric progression with the symbol d replacing
the normal r for the common ratio. For the given series d is actually 1 but this
formula becomes 0/0. Formula (v) is clearly the expression for the nth term, as
derived from the formula for the last (nth) term of an arithmetic series
l  a  (n  1)d 
C:
Correct.
Yes. Formula (iii) is the sum of n terms of an arithmetic series and formula (iv) is
derived from it and can be used in this form.
D:
Incorrect.
Formula (iii) is the sum of n terms of an arithmetic series but formula (v) is clearly
that for the nth term.
E:
Incorrect.
Formula (ii) is the sum to n for squares of the first n natural numbers,
i.e 12+22+32+... but formula (iv) is correct.
(i):
(ii):
(iii):
this is the formula for a geometric series with a =1, and the common ratio =1.
is the sum to n for squares of the first n natural numbers, i.e 12+22+32+...
is the formula for the sum of n terms of an arithmetic series and the given series is an arithmetic
series starting at unity, with a common difference of unity.
this is specifically the correct formula and derivable from (iii).
this is the nth term derived from the formula for the last term of an arithmetic series
l  a  (n  1)d  and in this case a =1, d=1.
(iv):
(v):
SEQUENCES AND SERIES Q 25
Try to do this problem without the use of a calculator
Using only the first 4 terms in the power series expansion of ex , calculate the value of e.
X
A:
1.732
X
B:
2.500
Y
C:
2.666
X
D:
2.708
X
E:
2.718
A:
Incorrect.
This is the value of the square root of 3. A useful number to know nevertheless!
The correct value and work is shown below:
B:
Incorrect.
You have used only 3 terms. Mind you, the error is only -8% from the
exact value for e of 2.718. The correct answer and work is shown below:
C:
Correct.
See the work below if you need to. The error between this and the exact value of e
is only about -2%.
D:
Incorrect.
This is the value using the first 5 terms. The error between this and the exact value
for e is now only -0.38%. I bet you used a calculator!!!
The correct answer and work is shown below:
E:
Incorrect.
This happens to be the exact value of e to 4 significant figures. Did you
remember it? The correct answer and work is shown below:
x 2 x3 x 4 x5


 ......
2! 3! 4! 5!
1 1
Putting x=1 gives, with only 4 terms used; e  1  1    1  1  0.5  0.166  2.666
2! 3!
The power series for the exponential e x  1  x 
SEQUENCES AND SERIES Q 26
The sum to infinity of some of the series below may converge. Identify which one if any do.
(i):
(iv):
1 1 1 1 1 1
      ...... (ii):
2 3 4 5 6 7
1 1 1 1 1 1
1        ...... (v):
2 3 4 5 6 7
1
1  2  3  4  5  ..... (iii):
1
9+7+5+3+1-1-3.....
1 1 1 1 1
    .....
1! 2! 3! 4! 5!
X
A:
(i) and (ii)
X
B:
(iii) and (iv)
X
C:
(v) and (i)
X
D:
(iii)
Y
E:
(iv) and (v)
A:
Incorrect.
The series (i) does not converge and (ii) is an arithmetic series and they never do.
See below for more details.
B:
Incorrect.
(iii) is an arithmetic series and they never converge, but (iv) does.
See below for more details.
C:
Incorrect.
This is of course the power series for e the exponential and therefore converges,
whereas (i) does not. See below for more details.
D:
Incorrect.
(iii) is an arithmetic series as is (ii) and they never converge.
See below for more details.
E:
Correct.
(iv) converges and (v) is the power series for e the exponential.
(i):
This series can be regrouped thus: 1            ...... The first term in brackets
2 3 4 5 6 7 8
equals 7/12 with is >1/2, as is the second term in brackets. Therefore this series cannot converge.
(ii)
Arithmetic series never converge.
(iii)
Even this one!!!
(iv)
The series converges even though (i) did not! I guess you have to just know that!
(v)
This is the power series for e the exponential. Try to memorise it!
1 1
1 1
1
1
1
SEQUENCES AND SERIES Q 27
Which of the following infinite power series converge and are equal to each other?
(i):
   1   1
  1   1
  1   1
 
 
 ....... (ii): 1   
 
 ....... (iii): 1    
 
 ......
4  4  3!  4  5!
4  4  2!  4  3!
 4  2!  4  4!

3
5
2

(iv): 1 

   
       ......
4 4 4
2
3
X
A:
(v) and (iv)
X
B:
(iii) and (ii)
X
C:
(i) and (v)
X
D:
(iv) and (iii)
Y
E:
(ii) and (i)
A:
Incorrect.
 
 
4 4
(v):
2
 
1  
4
4
3
2
3
1   1
 
 .......
3!  4  5!
5
1   1
 
 .......
2!  4  4!
4
These are the power series for tan

4
and a geometric series with a  1, r 

4
See below for explanation:

B:
Incorrect.
These are the power series for e 4 and cos

4
See below for explanation:
C:
Incorrect.
These are the power series for sin

4
and tan

4
See below for explanation:
D:
Incorrect.
These are a geometric series with a  1, r 

4

e4
and the power series for
See below for explanation:
E:
Correct.
(i)
This is the power series for sin
(iii)
(v)
These are the power series for sin
This is the power series for

4

e4
This is the power series for tan

4
and cos

4
which converge and are equal.

(ii)
This is the power series for cos
(iv)
This is a geometric series with a  1, r 

4
(i), (ii), (iii) and (v) definitely converge so the correct answer is E
4

4
SEQUENCES AND SERIES Q 28
Without using a calculator
Calculate the 10th term in the sequence 16, 32, 64, ........
X
A:
1024
X
B:
2048
X
C:
4094
X
D:
8190
Y
E:
none of these
A:
Incorrect.
The series is a geometric series with a first term a =16 and a common ratio r=2.
The nth term is ar n 1  16  29  24 29  213  8192 . Use your fingers!
B:
Incorrect.
The series is a geometric series with a first term a =16 and a common ratio r=2.
The nth term is ar n 1  16  29  24 29  213  8192 . Use your fingers!
C:
Incorrect.
The series is a geometric series with a first term a =16 and a common ratio r=2.
The nth term is ar n 1  16  29  24 29  213  8192 . Use your fingers!
D:
Incorrect.
The series is a geometric series with a first term a =16 and a common ratio r=2.
The nth term is ar n 1  16  29  24 29  213  8192 . Use your fingers!
E:
Correct.
Yes, this time none of the answers are right. Ha ha ha! Or perhaps: WELL DONE
SEQUENCES AND SERIES Q 29
The 10th term in an arithmetic series is 150 and the 12th term is 170. The first term is therefore:
X
A:
40
X
B:
50
Y
C:
60
X
D:
70
X
E:
80
A:
Incorrect.
Don't guess! It is quite difficult to get this wrong if you use the formula.
See below:
B:
Incorrect.
Don't guess! It is quite difficult to get this wrong if you use the formula.
See below:
C:
Correct.
A bit too easy? Did you use the formula for the nth term or just spot the
common difference?
D:
Incorrect.
Don't guess! It is quite difficult to get this wrong if you use the formula.
See below:
E:
Incorrect.
Don't guess! It is quite difficult to get this wrong if you use the formula.
See below:
Using the formula for the nth term l  a  (n  1)d  and putting in the given numbers:
150  a  9d  and 170  a  11d  gives d=10 and a =60
SEQUENCES AND SERIES Q 30
An arithmetic series starting at 8 has a sum to10 terms of 215. What is its common difference?
X
A:
+0.55
X
B:
-3
X
C:
+3.5
X
D:
+2.7
Y
E:
+3
A:
Incorrect.
I think you have made two mistakes in the formula for the sum of an arithmetic
series. Firstly you have used n instead of n/2 and secondly used n rather
than (n-1). See below for correct working:
B:
Incorrect.
I think you just made a sign mistake at the end. Bad luck, but it is a silly mistake
since the common difference clearly cannot be negative.
See below for correct working:
C:
Incorrect.
You have probably made two mistakes in the formula for the sum of an arithmetic
series. Firstly you put a rather than 2a and secondly used n not (n-1).
See below for correct working:
D:
Incorrect.
You have used n not (n-1) in the formula for the sum.
See below for correct working:
E:
Correct.
Not too difficult. See below for my method:
n
2a  (n  1)d . Inserting the values given:
2
S10  516  9d   215 16  9d  43 9d  27, giving d  3
The sum to n terms for an arithmetic series is Sn 