Geometric Progessions (C)

If r is a complex number and N is a positive integer, then the sum

$$\sum_{k=0}^{N-1} r^k \equiv 1 + r + \cdots + r^{N-1}$$

is called a geometric progression with N terms. It is easy to verify by long division that if $r\neq 1$ then

$$\sum_{k=0}^{N-1} r^k = \frac{r^N-1}{r-1} = \frac{1-r^N}{1-r}.$$

Of course, if r= 1 we have

$$\sum_{k=0}^{N-1} r^k = N.$$

Geometric progressions often arise as the soution of factoring problems involving differences of powers:

$$\frac{x^4-16}{x-2} = 8\frac{\left(\frac{x}{2}\right)^4-1}{\frac{x}{2}-1}$$

and by letting r = x/2 we see that

$$\begin{array} {rcl} \displaystyle{\frac{x^4-16}{x-2}} & = & ... ... \frac{x}{2} + 1\right)}\ & = & x^3 + 2x^2 + 4x + 8\end{array}$$

The general formula for N an positive integer and z and a distinct complex numbers is that

$$\frac{z^N-a^N}{z-a} = \sum_{k=0}^{N-1}z^ka^{N-1-k},$$

which is easily verified by the same technique as in the example. Well-known special cases are

$$\frac{z^2-a^2}{z-a} = z + a\;\;{\rm or}\;\; z^2-a^2 = (z-a)(z+a);$$

$$\frac{z^3-a^3}{z-a} = z^2 + az + a^2\;\;{\rm or}\;\; z^3-a^3 = (z-a)(z^2+az+z^2);$$

and

$$\frac{z^3+a^3}{z+a} = \frac{z^3-(-a)^3}{z-(-a)} = z^2 - az + a^2\;\;{\rm or}\;\; z^3+a^3 = (z+a)(z^2-az+z^2).$$

Exercises

1.

Find $${\sum_{k=0}^{10} 2^k}$$.

2.

Find $${\sum_{k=0}^{11} \left(\frac{1}{2}\right)^k}$$.

3.

Find $${\sum_{k=0}^{10} 2^{-k}}$$.

4.

Find $${\sum_{k=3}^{11} \left(\frac{-1}{2}\right)^k}$$.

5.

If x⁶ - 64 is divided by x-2, what is the result?

6.

If x - 2 is divided by $\sqrt[3]{x}-\sqrt[3]{2}$, what is the result?