This is a geometric series.  Use the following formula, where $\displaystyle a$ is the first term of the series, and $\displaystyle r$ is the ratio that must be less than 1.  If $\displaystyle r$ is greater than 1, the series diverges.

$\displaystyle \frac{a}{1-r} = \frac{1}{1-\frac{\sqrt2}{2}}= \frac{1}{\frac{2-\sqrt2}{2}} = \frac{2}{2-\sqrt2}$

Rationalize the denominator.

$\displaystyle \frac{2}{2-\sqrt2} \cdot \frac{2+\sqrt2}{2+\sqrt2} = \frac{2(2+\sqrt2)}{4-2}= 2+\sqrt2$

The problem can be reconverted using a summation symbol, and it can be seen that this is geometric.

$\displaystyle 1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...=\sum_{n=0}^{\infty} \left ( \frac{1}{4} \right )^n$

Since the ratio is less than 1, this series will converge.  The formula for geometric series is:

$\displaystyle \frac{a}{1-r}$

where $\displaystyle a$ is the first term, and $\displaystyle r$ is the common ratio.  Substitute these values and solve.

$\displaystyle \frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

The sum must be alternating, and after one period you should have the worm at 1m. After two periods, the worm should be at 2/3m. There is only one sum for which that is true.

First, we reduce the series into a simpler form.

$\displaystyle \sum_{n=0}^{\infty}(2)^{-2n}= \sum_{n=0}^{\infty}(\frac{1}{2})^{2n}= \sum_{n=0}^{\infty}(\frac{1}{4})^{n}$

We know this series converges because

$\displaystyle \lim_{n \rightarrow \infty}(\frac{1}{4})^{n}=0$

By the Geometric Series Theorem, the sum of this series is given by

$\displaystyle \sum_{n=0}^{\infty}(\frac{1}{4})^{n}= \frac{(\frac{1}{4})^0}{1-\frac{1}{4}}= \frac{4}{3}$

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$\displaystyle S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$, where n is the number of terms to sum up, r is the common ratio, and $\displaystyle a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$\displaystyle S_{15}=\frac{a_{1}*(1-r^n)}{1-r}$

$\displaystyle S_{15}=\frac{4*(1-(\frac{1}{3})^{15})}{1-\frac{1}{3}}$

$\displaystyle S_{15}=\frac{4*3(1-\frac{1}{3^{15}})}{2}$

$\displaystyle S_{15}=6*(1-\frac{1}{3^{15}})$

$\displaystyle S_{15}=5.999...$

Rounding, $\displaystyle S_{15}=6$

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$\displaystyle S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$, where n is the number of terms to sum up, r is the common ratio, and $\displaystyle a_{1}$ is the value of the first term.

We have $\displaystyle a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$\displaystyle a_{1}=6$

$\displaystyle r=\frac{a_{2}}{a_{1}}$ 

$\displaystyle r=\frac{12.6}{6} = 2.1$

$\displaystyle S_{16}=\frac{a_{1}*(1-r^n)}{1-r}$

$\displaystyle S_{16}=\frac{6*(1-2.1^{16})}{1-2.1}$

$\displaystyle S_{16}=780305$

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$\displaystyle S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$, where n is the number of terms to sum up, r is the common ratio, and $\displaystyle a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$\displaystyle S_{10}=\frac{a_{1}*(1-r^n)}{1-r}$

$\displaystyle S_{10}=\frac{11*(1-(\frac{2}{20})^{10})}{1-\frac{2}{20}}$

$\displaystyle S_{10}=\frac{11*(1-(\frac{1}{10})^{10}}{\frac{9}{10}} = \frac{11*10*(1-\frac{1}{10^{10}})}{9}$

$\displaystyle S_{10}=\frac{110}{9}(1-\frac{1}{10^{10}})$

$\displaystyle S_{10}=12.2222...$

Rounding, $\displaystyle S_{10}=12$