# Applying rules of algebra when working with multiplication and exponents

I'm taking an online course and help is hard to find. This specific problem has to do with recurrence relation. I apologize for being too general but I'm just looking for help in how to go about solving this problem.

The entire problem is:

Let $$d_0,d_1,d_2,\ldots$$ be defined by the formula $$d_n=3^n-2^n$$ for all integers $$n\ge0$$. Show that this sequence satisfies the recurrence relation: $$d_k=5d_{k-1}-6d_{k-2}$$.

The step I can do without any trouble is finding the statements that represent the values we're dealing with in the relation, $$d_{k-1}$$ and $$d_{k-2}$$:

$$d_k=3^k-2^k$$

$$d_{k-1}=3^{k-1}-2^{k-1}$$

$$d_{k-2}=3^{k-2}-2^{k-2}$$

However, when it comes time to plugging that into the relation and simplifying it down to the original definition of $$d_k=3^k-2^k$$ I fail miserably.

$$d_k=5d_{k-1}-6d_{k-2}$$

$$=5\left(3^{k-1}-2^{k-1}\right)-6\left(3^{k-2}-2^{k-2}\right)$$

$$=5\cdot3^{k-1}-5\cdot2^{k-1}-6\cdot3^{k-2}+6\cdot2^{k-2}$$

$$=5\cdot\frac{3^k}{3}-6\cdot\frac{3^k}{3^2}-5\cdot\frac{2^k}{2}+6\cdot\frac{2^k}{2^2}$$

Is that a decent start? Is there a better way to go, like breaking everything down into their simplest components, like so:

$$d_k=5d_{k-1}-6d_{k-2}$$

$$=(2+3)\left(3^{k-1}-2^{k-1}\right)-(2\cdot3)\left(3^{k-2}-2^{k-2}\right)$$

$$=2\cdot\frac{3^k}{3}-2\cdot\frac{2^k}{2}+3\cdot\frac{3^k}{3}-3\cdot\frac{2^k}{2}-\left(\left(2\cdot\frac{3^k}{3^2}-2\cdot\frac{2^k}{2^2}\right)-\left(3\cdot\frac{3^k}{3^2}-3\cdot\frac{2^k}{2^2}\right)\right)$$

But then what?

If I go either of these routes, I get stuck. I feel I have two issues: a.) identifying the way to go that seem the most logical; and, b.) working towards a solution. I don't know if I missed a big chunk in Algebra or if my brain just doesn't see what's going on.

What am I missing? Are either of these steps valid things to try? What are some general rules to follow to work these out?

Also, what specific discipline of Algebra is this? I don't think my course is introducing us to this stuff. I think it assumes we already know how to work these out.

• Have you tried to factor $3^{k}$ and $2^{k}$? – IEDC PHY Nov 3 '18 at 0:15
• While I appreciate your comment, I wasn't sure which line above to attempt to factor them out. – harperville Nov 3 '18 at 1:16

I start from what you wrote down: $$5d_{k-1}-6d_{k-2}=5\cdot3^{k-1}-6\cdot3^{k-2}-5\cdot2^{k-1}+6\cdot2^{k-2}$$ and I want to get to $$d_k=3^k-2^k$$.
I rewrite $$3^{k-1}=3\cdot3^{k-2}$$, and similarly for $$2^{k-1}$$: $$5\cdot3\cdot3^{k-2}-6\cdot3^{k-2}-5\cdot2\cdot2^{k-2}+6\cdot2^{k-2}$$ $$=15\cdot3^{k-2}-6\cdot3^{k-2}-10\cdot2^{k-2}+6\cdot2^{k-2}$$ I collect $$3^{k-2}$$ and $$2^{k-2}$$: $$=(15-6)3^{k-2}-(10-6)2^{k-2}=9\cdot3^{k-2}-4\cdot2^{k-2}=3^2\cdot3^{k-2}-2^2\cdot2^{k-2}$$ From here I collapse the exponents to get the desired result. $$=3^k-2^k$$