Questions: Find the geometric sum: 6+12+24+...+6,144.

Solution Steps

The given series is \(6 + 12 + 24 + \ldots + 6,144\). This is a geometric series because each term is obtained by multiplying the previous term by a constant ratio.

The first term \(a\) of the series is \(6\). To find the common ratio \(r\), divide the second term by the first term: \[ r = \frac{12}{6} = 2. \] Thus, the common ratio \(r = 2\).

The last term of the series is \(6,144\). The general formula for the \(n\)-th term of a geometric series is: \[ a_n = a \cdot r^{n-1}. \] Substitute the known values: \[ 6,144 = 6 \cdot 2^{n-1}. \] Divide both sides by \(6\): \[ 2^{n-1} = \frac{6,144}{6} = 1,024. \] Express \(1,024\) as a power of \(2\): \[ 1,024 = 2^{10}. \] Thus: \[ 2^{n-1} = 2^{10} \implies n-1 = 10 \implies n = 11. \] There are \(11\) terms in the series.

The sum \(S_n\) of the first \(n\) terms of a geometric series is given by: \[ S_n = a \cdot \frac{r^n - 1}{r - 1}. \] Substitute \(a = 6\), \(r = 2\), and \(n = 11\): \[ S_{11} = 6 \cdot \frac{2^{11} - 1}{2 - 1} = 6 \cdot (2^{11} - 1). \] Calculate \(2^{11}\): \[ 2^{11} = 2,048. \] Thus: \[ S_{11} = 6 \cdot (2,048 - 1) = 6 \cdot 2,047 = 12,282. \]

Final Answer