Recurrence relation question (generating direct formula)

Question

Solve this recurrence relation by generating its direct formula: $$a_n = 3a_{n-1} + 2n, a_0 = 1$$ Using the direct formula find the $10th$ term of this recurrence relation.

My answer:

$a_{10} = 147601$

Is this correct?

Answer 1

You have not produced the required direct formula.

Hint

Note that $a_{n} + n = 3 a_{n-1} + 3 n = 3(a_{n-1} + n) = 3(a_{n-1} + n-1) + 3$.

Answer 2

I'm not really sure but after I solve it my recurrence relationship is 3^n + 26n. so I sub n = 10 in and I obtain 59309.

Answer 3

Almost correct. Note: $$a_{10}=3a_9+2\cdot 10=3\cdot 49197+20=147611.$$ However, can you find $a_{10}$ directly, that is without finding $a_1,a_2,...,a_9$? You will need a formula for it. There are several methods to find the general formula: $$a_n=\frac52 \cdot 3^n-n-\frac32 \Rightarrow a_{10}=\frac52\cdot 3^{10}-10-\frac32=147611.$$