Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading the solution of a recurrence relation exercise. I think I understand the solution, except for the two following steps:

$$\cdots = \displaystyle{\underbrace{\left( -1\right) ^{n}\sum _{k=0}^{n}\left( -2\right) ^{k}}_{(1)} =\underbrace{\left( -1\right) ^{n}\dfrac {\left( -2\right) ^{k}} {-3}\bigg|_{0}^{n+1}}_{(2)}} = \underbrace{\frac {(-1)^{n+1}}{3}((-2)^{n+1}-1)}_{(3)}= \cdots$$

How do you go from $(1)$ to $(2)$ - which, as we see, converts the summation in $(1)$ to a closed form in $(2)$, but that still uses the $k$ index - and then from $(2)$ to $(3)$ - thereby losing the index $k$?

share|improve this question

3 Answers 3

up vote 2 down vote accepted

Going from (1) to (2), they are using $1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r}$ with $r=-2$, and then going from (2) to (3) they are substituting $n+1$ and $0$ in for $k$ and subtracting, and using that $\frac{(-1)^{n}}{-1}=(-1)^{n+1}$.

share|improve this answer

$$ \sum_{k=0}^n x^k = \frac{x^{n+1} - 1}{x-1} $$

share|improve this answer
1  
$$f(x) |_b^a = f(a) -f(b)$$ –  what'sup Sep 6 '13 at 22:37

For getting to step (2), let $S=\sum\limits_{k=0}^n(-2)^k$. Observe multiplying throughout $-2$ yields $-2S=\sum\limits_{k=1}^{n+1}(-2)^k$. The difference between the two is then just:$$\begin{align*}-2S-S&=\sum_{k=1}^{n+1}(-2)^k-\sum_{k=0}^n(-2)^k\\-3S&=(-2+4-8+\dots+(-2)^n+\color{red}{(-2)^{n+1}})-(\color{red}1-2+4-8+\dots+(-2)^n)\\-3S&=\color{red}{(-2)^{n+1}-1}+(2-2)+(4-4)+\dots\\-3S&=(-2)^{n+1}-1\\S&=\frac{(-2)^{n+1}}{-3}-\frac1{-3}=\left[\frac{(-2)^k}{-3}\right]_{k=0}^{k=n+1}\text{ or }\frac{(-2)^k}{-3}\Biggr|_0^{n+1}\end{align*}$$ In general, the above argument shows that for $S=1+r+r^2+\dots+r^n$ we have $$S=\frac{r^{n+1}-1}{r-1}=\frac{r^k}{r-1}\Biggr|_0^{n+1}$$


For step (3), we merely expand:$$\frac{(-2)^k}{-3}\Biggr|_0^{n+1}=\frac{(-2)^{n+1}}{-3}-\frac{(-2)^0}{-3}=\frac{(-1)^{n+1} 2^{n+1}}{-3}-\frac1{-3}$$

In case you are unfamiliar with the notation, we write:$$F(k)\bigg|_a^b=F(b)-F(a)$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.