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

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq