Solve the following recurrence relations. a. x(n) = x(n − 1) + 5 for n > 1, x(1) = 0b. x(n) = 3x(n − 1) for n > 1, x(1) = 4c. x(n) = x(n − 1) + n for n > 0, x(0) = 0

(a)x(n)=5(n-1)

(b)x(n)=4 X 3ⁿ⁻¹

(c)x(n)=n(n+1)/2

Step-by-step explanation:

(a)x(n) = x(n − 1) + 5 for n > 1, x(1) = 0

Put n=n-1

x(n-1)=x(n-2)+5

Recall: x(n-1)= x(n)-5

Substitute:

x(n)-5=x(n-2)+5

x(n)=x(n-2)+5+5

Put n=n-2

x(n-2)=x(n-3)+5

Recall: x(n-2)= x(n)-5-5

Substitute:

x(n)-5-5=x(n-3)+5

x(n)=x(n-3)+5+5+5

Generalising

x(n)=x(n-i)+5i for i<n

Put i=n-1

x(n)=x(n-(n-1))+5(n-1)

x(n)=x(1)+5(n-1)=0+5(n-1)

x(n)=5(n-1)

(b)x(n) = 3x(n − 1) for n > 1, x(1) = 4

Put n=n-1

x(n-1)=3x(n-2)

Recall: x(n-1)= x(n)/3

Substitute:

x(n)/3=3x(n-2)

x(n)=3 X 3x(n-2)

Put n=n-2

x(n-2)=3x(n-3)

Recall: x(n-2)= x(n)/3²

Substitute:

x(n)/3²=3x(n-3)

x(n)= 3³x(n-3)

Generalising

x(n)=3ⁱ x(n-i) for i<n

Put i=n-1

x(n)=3ⁿ⁻¹ x(n-(n-1))

x(n)=3ⁿ⁻¹ x(1)

x(n)=3ⁿ⁻¹*4

x(n)=4 X 3ⁿ⁻¹

(c) x(n) = x(n − 1) + n for n > 0, x(0) =0

Put n=1

x(1)=x(1-1)+1

x(1)=0+1=1

Put n=2

x(2)=x(2-1)+2

x(2)=x(1)+2=1+2=3

Put n=3

x(3)=x(3-1)+3=x(2)+3=3+3=6

Generalising

x(n)=n(n+1)/2

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: