Finding the Sum of Geometric Series of n Terms being 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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