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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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Review exercises 6 223

15 (a) Write down the seriesgiven by ∑ 5

r=1 r 2 .

(b) The sumofthe squaresofthe firstnwhole

numbers can be foundfrom the formula

n∑

r 2 n(n +1)(2n +1)

=

6

r=1

Use thisformula to find

(i) ∑ 5

r=1 r 2 , (ii) ∑ 10

r=1 r 2 , (iii) ∑ 10

r=6 r 2 .

16 (a) Writedown the series given by ∑ 6

r=1 r 3 .

(b) The sumofthe cubesofthe firstnwhole

numbers canbe foundfrom the formula

(

n∑

r 3 =

n(n +1)

2

r=1

Use thisformula to find

) 2

(i) ∑ 6

r=1 r 3 , (ii) ∑ 12

r=1 r 3 , (iii) ∑ 12

r=7 r 3 .

17 Thethirdterm ofan arithmeticprogression is 18.The

fifthterm is28. Find the sum of20terms.

18 (a) Find an expression forthe general term in the

sequence

2,5,10,17,...

(b) Define the sequence in terms ofarecurrence

relation.

19 (a) Show thatthe equationx 3 +2x −14 = 0 canbe

rearrangedintothe formx = 3√ 14 −2x.With

x 0 = 2 use simple iteration to findarootofthe

equation.

(b) Rearrange the equation 0.8sinx −0.5x = 0 into

the formx =g(x).Withx 0 = 2 use simple

iteration to findarootofthe equation.

(c) Rearrange the equationx 3 = 2e −x intothe form

x =g(x).Withx 0 = 0usesimpleiterationtofind

a rootofthe equation.

20 Writeoutexplicitlythe firstfour terms ofthe series

∞∑

m=0

(−1) m x 2m

2 2m (m!) 2 .

21 Writeoutexplicitlythe firstfour terms ofthe series

∞∑

m=0

(−1) m x 2m+1

2 2m+1 m!(m +1)! .

Solutions

1 (a) 1,−1,1, −1,1 (b) 1, − 1 3 , 1 5 , −1 7 , 1 9

2 (a) 8k−7 k 1 (b) (−1) k k 1

4 1+5x+15x 2 +30x 3

(

+···)

1

5 1 −x+ 2x2

27 3 − 10x3

27

7 0

8

9 0

27

8

11 0, 2, 3, 3, −1, −7

12 (a) 0 (b) nolimit (c) nolimit

(d) a (e) nolimit

13 (a) − 40

9

(b) 19

14 2, 3 4

15 (a) 1+4+9+16+25

(b) (i) 55 (ii) 385 (iii) 330

16 (a) 1+8+27+64+125+216

(b) (i) 441 (ii) 6084 (iii) 5643

17 1110

18 (a) x[k]=k 2 +1,k=1,2,3,...

(b)x[k+1]=x[k]+2k+1

19 (a) 2.13

(b) x = 1.6sinx,1.6

(c)x= 3 √

2e −x ,0.93

20 1− 1 4 x2 + 1 64 x4 − 1

2304 x6 +···

21 1 2 x − 1 16 x3 + 1

384 x5 − 1

18432 x7 +···

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