Question: "Find a recurrence relations for the sequence {an..."

Mohamed Nazar

Answered question

2022-04-28

Find a recurrence relations for the sequence {an }
given by n n
an  A.2  B.(3) .

alenahelenash

Expert2023-05-02Added 556 answers

The sequence

{a_{n}}

is defined by the formula

a_{n} = A n^{2} + B (- 3)^{n}

. To find a recurrence relation for this sequence, we need to express each term in terms of previous terms.
Let's first find

a_{n + 1}

a_{n + 1} = A (n + 1)^{2} + B (- 3)^{n + 1} = A (n^{2} + 2 n + 1) - 3 B (- 3)^{n} = (A n^{2} + B (- 3)^{n}) + 2 A n + A - 3 B (- 3)^{n} = a_{n} + 2 A n + A - 3 B (- 3)^{n}

Therefore, we have the recurrence relation:

a_{n + 1} = a_{n} + 2 A n + A - 3 B (- 3)^{n}

Note that this recurrence relation depends on the values of

A

and

B

. To determine

A

and

B

, we need two initial conditions. Let's assume that

a_{0} = k_{1}

and

a_{1} = k_{2}

for some constants

k_{1}

and

k_{2}

. Then we have:

a_{0} = A (0)^{2} + B (- 3)^{0} = A

a_{1} = A (1)^{2} + B (- 3)^{1} = A - 3 B

Solving for

A

and

B

, we get:

A = a_{0}

B = \frac{A - a_{1}}{3}

Substituting these values into the recurrence relation, we get:

a_{n + 1} = a_{n} + 2 n a_{0} + a_{0} - \frac{(a_{1} - a_{0}) (- 3)^{n}}{3}

Therefore, the recurrence relation for the sequence

{a_{n}}

is:

a_{n + 1} = a_{n} + 2 n a_{0} + a_{0} - \frac{(a_{1} - a_{0}) (- 3)^{n}}{3}

with

a_{0}

and

a_{1}

given as initial conditions.

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