Yahoo Answers: Linear Homogeneous Recurrence - JunkYardDawg

[MarkFL]

Here is the question:

Recurrence relation help!?

For the recurrence relation where a_(n) = -2a_(n - 1) and a_0 = 10

Find:

A) An explicit formula in terms of n

B) The formula for the sum of the first n terms of the relation

C) The sum of the first 10 terms of the relation.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello JunkYardDawg,

A.) Let's rewrite the recursion as:

\(\displaystyle a_{n}+2a_{n-1}=0\)

From this, we can see the characteristic root is:

\(\displaystyle r=-2\)

And so the closed form is:

\(\displaystyle a_n=k(-2)^n\)

Now, using the given initial value, we may determine $k$:

\(\displaystyle a_0=k(-2)^0=k=10\)

Hence the closed form solution is:

\(\displaystyle a_n=10(-2)^n\)

B.) The sum of the first $n$ terms is:

\(\displaystyle S_{n}=10\sum_{k=0}^{n-1}\left((-2)^k \right)\)

If we multiply through by $-2$, then we may write:

\(\displaystyle -2S_{n}=10\sum_{k=0}^{n-1}\left((-2)^{k+1} \right)=S_{n}-10\left(1-(-2)^n \right)\)

This implies:

\(\displaystyle 3S_{n}=10\left(1-(-2)^n \right)\)

Divide through by $3$:

\(\displaystyle S_{n}=\frac{10}{3}\left(1-(-2)^n \right)\)

C.) Using the formula from part B), we find:

\(\displaystyle S_{10}=\frac{10}{3}\left(1-(-2)^{10} \right)=-\frac{10230}{3}=-3410\)