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Archangel Macsika

Determine values of the constants A and B such that an = An + B is a solution of recurrence relation an = 2an−1 + n + 5. Hence, find the solution of this recurrence relation with a0 = 4.

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Let us determine values of the constants A and B such that a_n = An + B is a particular solution of the recurrence relation a_n = 2a_{n−1} + n + 5.

It follows that

An+B=2(A(n-1)+B)+n+5=2An-2A+2B+n+5

=(2A+1)n+(2B-2A+5),

and thus

A=2A+1 and B=2B-2A+5.

Therefore, A=-1 and B=2A-5=-7.

We conclude that a_n=-n-7 is a paticular solution of the recurrence relation.


Further, let us find the general solution of this recurrence relation with a_0 = 4.

It follows that the characteristic equation k-2=0 has the solution k=2, and hence the general solution is of the form a_n=C\cdot2^n-n-7. Since 4=a_0=C-7, we conclude that C=11.

Consequently, the general solution of this recurrence relation a_n = 2a_{n−1} + n + 5 with a_0 = 4 is

a_n=11\cdot2^n-n-7.


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