等比数列の和

等比数列

\begin{align}
S_n = \{a,ar,ar^2,\cdots,ar^n-1\}
\end{align}

の和は

\begin{align}
S_n &= \{a,ar,ar^2,\cdots,ar^n-1\}\\
rS_n &= \{ar,ar^2,ar^3,\cdots,ar^n\}
\end{align}

の差

\begin{align}
S_n -rS_n= a+ar+ar^2+\cdots+ar^n-1 -(ar+ar^2+ar^3+\cdots,ar^n\})
\end{align}

より

\begin{align}
(1-r) S_n = a ( 1 – r^n )
\end{align}

従って

\begin{align}
S_n = \frac{a ( 1 – r^n )}{(1-r)}= \frac{a ( r^n – 1 )}{(r-1)}
\end{align}

を得る。