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