The Ultimate Guide to Geometric Series Sums: 1 to 2

Understanding Geometric Series

A geometric series is a sequence of numbers where each term is obtained by multiplying the previous term by a constant, known as the common ratio. When a geometric series of n terms is inserted between 1 and 2, it creates a unique sum that can be calculated using specific formulas and techniques.

The Sum Formula

The sum of a geometric series of n terms inserted between 1 and 2 can be found using the following formula:

$S_n = 1 + r + r^2 + r^3 + \cdots + r^{n-1} + 2$

where $S_n$ is the sum of the series, $r$ is the common ratio, and $n$ is the number of terms.

Deriving the Common Ratio

To use the sum formula, you must first determine the common ratio, $r$. This can be done by considering the relationship between the first and last terms of the series, which are always 1 and 2, respectively.

$\displaystyle r^{n-1} = \frac{2-1}{1} = 1$

Solving this equation for $r$ will give you the common ratio in terms of $n$.

Simplifying the Sum

Once you have the common ratio, you can substitute it into the sum formula and simplify the expression. This will give you a general formula for the sum of a geometric series of n terms inserted between 1 and 2.

By mastering these concepts and techniques, you’ll be able to efficiently calculate the sum of any geometric series of n terms inserted between 1 and 2.

Suppose $n$ consecutive geometric terms are inserted between $1$ and $2$. Write the sum of these $n$ terms in terms of $n$.

$$\Large \underbrace{1, \overbrace{u_1, u_2, u_3, \cdots, u_{n}}^{S_n}, 2}_{S_{n+2}}$$

\require{AMSsymbols} \displaystyle \begin{align} S_{n+2} &= 1+u_1 + u_2 + u_3 + \cdots + u_n + 2 \\ &= 1 + S_n + 2 \\ S_n &= S_{n+2}-3 \color{red}{\cdots (1)} \\ S_{n+2} &= \frac{a(r^{n+2}-1)}{r-1} &\color{green}{a \text{ is the first term, and } r \text{ is the common ratio}} \\ &= \frac{r^{n+2}-1}{r-1} \color{red}{\cdots (2)} &\color{green}{a=1} \\ u_1 &= r &\color{green}{1 \times r = u_1} \\ u_2 &= r^2 \\ u_3 &= r^3 \\ &\cdots \\ u_n &= r^{n} \\ 2 &= r^{n+1} &\color{green}{u_n \times r = 2} \\ r &= 2^{\frac{1}{n+1}} \\ S_{n+2} &= \frac{\left(2^{\frac{1}{n+1}}\right)^{n+2}-1}{2^{\frac{1}{n+1}}-1} &\color{red}{\text{by } (2)} \\ &= \frac{2^{\frac{n+2}{n+1}}-1}{2^{\frac{1}{n+1}}-1} \\ \therefore S_n &= \frac{2^{\frac{n+2}{n+1}}-1}{2^{\frac{1}{n+1}}-1}-3 &\color{red}{\text{by } (1)} \end{align}

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