Ex 1e Solutions

Solutions to Exercise 1E CASIO ClassPad: 1. Open the Sequence application H 2. Input the following:

1 t n +1 = 3t n −1,

t1 = 6

t2 = 3t1 – 1 = 3 × 6 – 1 = 17

t3 = 3t2 – 1 = 3 × 17 – 1 = 50

Using CAS, TI-Nspire: 1. Set to AUTO or Approximate mode. 2. Open the Graphs & Geometry application and change the Graph Type to Sequence. 3. Input the following:

then press E 3. Tap the empty box next to an+1 so that a tick appears. 4. Tap # to display a table of values. 5. Tap 8 and change the table start value to 0 and the table end value to 10. 6. Scroll down to until you reach n = 7. then press · 4. Press /T to display a table of values for this sequence. 5. Scroll down to the cell that corresponds to n = 8.

∴t 8 =12029

∴t 8 =12029

CASIO ClassPad: 1. Refer to the instructions given in question one.

2 y n +1 = 2 y n + 6,

y2 = 2y1 + 6 = 2 × 5 + 6 = 16

y1 = 5

y3 = 2y2 + 6 = 2 × 16 + 6 = 38

Using CAS,

TI-Nspire: 1. In a new sequence plot input the following then press ·

2. Scroll down until you reach n = 9.

2. Set an appropriate Window Setting.

3. Press /T for a table of values. 4. Scroll down to the cell that corresponds to n = 10.

∴y10 = 5626

∴y10 = 5626 3. Tap $ to view the graph. 4. Tap 6 and set an appropriate Window Setting.

3 t t2 t3 t4 t5

= = = = =

1 1 t2 + t1 = 2 t3 + t2 = 3 t4 + t3 = 5

t6 t7 t8 t9 t10

= = = = =

t5 t6 t7 t8 t9

+ + + + +

t4 t5 t6 t7 t8

= = = = =

8 13 21 34 55

c S∞ – Sn = – = , 4

a = x + 5 = 25, r = 5 , n = 10 ∴ S∞ – S10

= = 5 × 25 ×

The first ten terms are: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55

= 4 a = 3, d = 4, n = 10 Sn = n2 [2a + (n – 1)d] ∴ S10 = 10 2 [2 × 3 + (10 – 1) × 4] = 5[6 + 9 × 4] = 5 × 42 = 210 5 S∞ = , a = 1, r = =

1

3 3 = 4

a = 6, r = –3, n = 8 ∴ S8 = = ((–3)8 – 1) = –9840 8 S∞ = , a = a, r =

×

2 + 1 2 + 1 = a(2 + 2) 1

6 a

= a (2 + )

x = x + 5

∴ x2 = = ∴ x =

(x + 5)(x – 4) x2 + x – 20 20

b x r = x + 5 20 = 25 4

= 5

410 57

7 Sn = ,

= = =

= 4

4

( 5 )10

a ÷ a = 2

1 2

9 a Sn = , n = 10, a = 1, r = x

10 a S∞ = , a = 1, r = sin θ =

2

∴ S10

= = (( x )10 – 1)

b

2



3  4   10

S10 = 41 −   

When x = 1.5,

 

b i. S∞ = , a = 1, r = x 2

= , x ≠ 2 = Now –1 < r < 1 x

∴ –1 < 2 < 1 ∴ –2 < x < 2 The infinite sum exists for –2 < x < 2 ii. Let S = , x ≠ 2 Given S = 2S10, = (( x )10 – 1)

2 x 10 1 ∴ (2 ) – 1 = – 2 x 1 ∴ ( 2 )10 = 2 x 1 1 ∴ 2 = ±( 2 ) 10

∴ x = ±2 × 2 =

9

± 2 10

= 2 ∴ 2(1 – sin θ ) = 1 1

∴ 1 – sin θ = 2 1 ∴ sin θ = 2 π

5π

π

5π

π

∴ θ = 6 , 6 , 6 ± 2π, 6 ± 2π, 6 ± 4π, … π

5π

∴ θ = 6 + 2kπ, 6 + 2kπ, k ∈ Z