2010 msm recurrence relation solutions)

H2 Maths (9740) 2010 MSM : RECURRENCE RELATION

1 A sequence of real numbers satisfies the recurrence relation

, .

for all integers and where k is a real number.

(i) When k = 1, evaluate . [1]

(ii) When k = , describe the behaviour of the sequence. [1]

(iii) If the sequence converges, determine algebraically the limit of the sequence.[2009/RJC/Promo/1]

[Solution](i) From the GC, = 1597.

(ii) From the GC, the sequence is increasing and converges to .

(iii) If the sequence converges, we have when .

Hence from , we have

………[1]

Squaring both sides we obtain

Since l < 0 from [1], we must have .

Alternatively, from (i) when , the sequence is not constant, thus we reject

2 A sequence is such that and

, for

(i) Express in the form , where A is a constant to be determined.

(ii) Determine whether the sequence converges. [2009/IJC/Promo/8]

[Solution]

1


(i)

=

So, A = 13.

(ii)

Therefore, the sequence converges.

2


3 a) In the figure below, a square is inscribed in a circle, a smaller circle is inscribed in the square and the process goes on indefinitely. Given that the radius of the largest circle is r, find the sum of the areas of all the circles in terms of r. [4]

b) The sequence of numbers x1, x2, x3,… is such that x1 = and for

all .

i) Write down the exact values of x2, x3 and x4. [2]

ii) Conjecture a possible expression for xn in terms of n. [1][2009VJC/JCT/4]

[Solution]

3


4 A sequence of positive integers u1, u2, u3, … is defined by u1 = 3 and

for .

(i) Using a graphic calculator, find un for n =1, 2, 3, 4. [2]

(ii) By comparing the values of with , or otherwise, write down a conjecture

for un in terms of n, where . [2](iii) Prove your conjecture using mathematical induction. [4](iv) State, giving a reason, whether the sequence converges or diverges as [2]

[2009/DHS/JC1 JCT/9] [Solution]

(i)

(ii) Conjecture :

(iii) Let P(n) be the proposition

When n =1, LHS of P(1) = 3 (given)

RHS of P(1) = P(1) is true.

Assume P( k ) is true for some

When n=k+ 1

RHS of P(k+1) =

LHS of P(k+1) =

P(k) is true P(k+1) is true

4


Since P(1) is true and P(k) is true P(k+1) is true, by mathematical induction, P(n) is true for

(iv)

As

ie. The sequence is covergent and converges to 0.

5 A recurrence relation is defined by , where n = 1, 2, 3, …

(i) If xn converges to when n is large, find the exact values of . [3]

(ii) If x1 = 0.5, obtain the first 4 terms of the sequence {xn}. What can you say about the sequence {xn} when n is large ? [2]

(iii) By sketching a suitable diagram, explain clearly that if xk > for some integer k , then xk + 1 < xk . [3]

[2009/TJC/Promo/10][Solution]

(i) When n is large, xn and xn+1 both converge tend to . Thus

3 – 5 = 0 (2 – 5 ) = 0. So = 0 or

(ii) By using a GC, x1 = 0.5 , x2 = or 1.86 , x3 = 2.31 and x4 = 2.22

The sequence {xn} converges to when n is large. (iii) Sketch the graph of

as shown below

or sketch

.

5


When x = 0, and , y = 0.

From the graph, whenever, xk > , y < 0 that is

6 (i) Find the range of values of k for which the equation

has real roots.

(ii) A sequence of real numbers satisfies the recurrence relation

for , where p is a non-zero constant.

Prove algebraically that, if the sequence converges, .

(iii) Suppose the sequence converges to 1, find the value of p.

(iv) By considering , prove that

for ,

for .

[2009/MJC/JCT/P1/Q9modified]

[Solution]

(i) For to have real roots

6


(ii) Since the sequence converges, as ,

Hence

To solve for l, we need (1) to have real roots and

from (i) for (1) to have real roots, .

(iii) From (ii) if the sequence converges to 1 Substitute into (1)

Therefore

(iv) Considering ,

For , from (i) it can be infer that has no real roots.

Also since coefficient of is positive, hence

Therefore

(proved)

For , from (i) it can be infer that has no real roots.

Also since coefficient of is positive , hence

Therefore ( )

(proved)

7 The sequence of positive numbers satisfies the recurrence relation

for .

(i) As , find the exact value of . [2]

(ii) Show that if then . [2]

(iii) Use a calculator to determine the behavior of the sequence{ }when Hence state briefly how the results in (ii) relate to the behavior of the sequence. [2]

[2009/SRJC/ Prelim/1/14(b)]

7

(1)


[Solution]

(i)

When

or (rej as is positive)

(ii)

Since and

so since

Alternative solution (1):

Since and

also since ,

8


Alternative solution (2):

use of GC to sketch

From the graph, when ,

(iii) It decreases and converges to 3.

As x1 = 4 >( =3), from (ii)

Since from GC, , from (ii) again,

Continuing in this manner, we have

which means that the sequence is decreasing.

Also since for all , therefore the sequence converges to

eventually.

9


8 The function f is defined by

.

Solve the equation exactly.

The sequence satisfies the recurrence relation for all

with .

(i) Show that the limit L, if it exists, satisfies the equation .

(ii) The points A and B have coordinates and respectively. By

considering the equation of line AB, show that is the x-intercept of the

line AB.

(iii) Using a graphical argument, deduce that is a convergent sequence.

(iv) Deduce the exact value of L.

[2009/NYJC/Promo/Q10]

[Solution]

(i) As , . Thus

Thus the limit L satisfies the equation .(ii) Equation of line AB:

At the x-intercept, y= 0. Thus

10


Thus is the x-intercept of the line AB.(iii) Consider the graph of f(x)

From the graph, and using (ii), as , decreases and is bounded by

. Thus is a convergent sequence.

From (iii), is a decreasing sequence. Thus 22L .

9 A sequence is defined by the recurrence relation as follows:

(i) If , show by induction that where .

(ii) If determine .

(iii) If write down the first few terms of the sequence and hence

give a conjecture for in terms of n.

A Comprehensive Guide H2 Mathematics For A level Vol 1Solution

(i) Let be the statement “ where .”

When n = 1, LHS = .

RHS = LHS.

Therefore is true.

Assume is true for some ,

11

x22 22

( 4, 2)A

1x2x3x

f ( )y x


i.e.

[Want to show that ]

LHS =

Hence true is true.

Since is true and true is true, hence by Mathematical Induction,

is true for all .

(ii)

Since as

(iii)

12


where

13

2010 msm recurrence relation solutions)

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