Question

Find the value of a4 for the recurrence relation an=2an.1+3, with a0=6.

Answer 1

Answer:

When n=1, a1=2a0+3, Now a2=2a1+3. By substitution, we get a2=2(2a0+3)+3.

Step-by-step explanation:

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Answer 2

The value of a₄ for the recurrence relation aₙ= 2aₙ₋₁+3 is 141.

Given:

The recurrence relation:

aₙ= 2aₙ₋₁+3

a₀= 6

To Find:

the value of a₄.

Solution:

The recurrence relation:

aₙ= 2aₙ₋₁+3     _(1)

Putting value of n=1 in equation (1)

a₁= 2a₁₋₁+3

a₁=2a₀+3

Putting value of a₀ in above equation

a₁= 2(6)+3= 15

Putting n=2 in equation (1)

a₂= 2a₂₋₁+3

a₂= 2a₁+3

Putting value of a₁ in equation (1)

a₂= 2(15)+ 3= 33

a₂= 33

Putting n= 3 in equation (1)

a₃= 2a₃₋₁+ 3

a₃= 2a₂+3

Putting a₂=33 in above equation

a₃= 2(33)+3

a₃= 66+3

a₃= 69

Putting n=4 in equation (1)

a₄= 2a₄₋₁+3

a₄= 2a₃+3

Putting value of a₃ in above equation

a₄= 2(69)+3

a₄= 138+3

a₄= 141

Hence, the value of a₄ for the recurrence relation aₙ= 2aₙ₋₁+3 is 141.

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