## College Algebra (10th Edition)

The sequence is geometric. $r=\sqrt3$ $S_{50}\approx 2,004,706,374,277$
$\bf\text{RECALL:}$ $\bf\text{(1) Arithmetic Sequence }$ A sequence is arithmetic if there exists a common difference $d$ among consecutive terms. $d=a_n-a_{n-1}$ The sum of the first $n$ terms of an arithmetic sequence is given by the formulas: $S_n=\frac{n}{2}(a_1 +a_n)$ or $S_n=\frac{n}{2}\left(2 a_1 + (n-1)d\right)$ $\bf\text{(2) Geometric Sequence }$ A sequence is geometric if there exists a common ratio $r$ among consecutive terms. $r=\dfrac{a_n}{a_{n-1}}$ The sum of the first $n$ terms of a geometric sequence is given by the formula: $S_{n}=a_1 \cdot \dfrac{1-r^n}{1-r}$ In the formulas listed above, $d$ = common difference $r$ = common ratio $a_1$ = first term $a_n$ = nth term $n$ = number of terms $\bf\text{List the first few terms of the sequence.}$ $\bf\text{Identify the sequence as arithmetic or geometric.}$ Substitute $1, 2, 3, 4$ for $n$ to list the first three terms: $a_1 =3^{1/2}=\sqrt3$ $a_2 = 3^{2/2}=3^1=3$ $a_3 = 3^{3/2}=\sqrt{3^3} = \sqrt{3^3(3)} = 3\sqrt3$ $a_4 = 3^{4/2}=3^2 = 9$ Note that the terms have no common difference. Thus, the sequence is not arithmetic. To check if the sequence is geometric, solve for the ratios of each pair of consecutive terms: $\dfrac{a_2}{a_1}=\dfrac{3}{\sqrt3} = \dfrac{3\cdot \sqrt3}{\sqrt3\cdot \sqrt3}=\dfrac{3\sqrt3}{3} = \sqrt3$ $\dfrac{a_3}{a_2}=\dfrac{3\sqrt3}{3} = \sqrt3$ $\dfrac{a_4}{a_3} = \dfrac{9}{3\sqrt{3}} = \dfrac{9 \cdot \sqrt3}{3\sqrt{3} \cdot \sqrt3} = \dfrac{9\sqrt3}{3 \cdot 3}=\dfrac{9\sqrt3}{9}=\sqrt3$ The sequence has a common ratio of $\sqrt3$, so it is geometric. $\bf\text{Find the sum of the first 50 terms}:$ With $a_1=\sqrt3$ and $d=\sqrt3$, solve for the sum of the first 50 terms using the formula in (1) above to obtain: $S_n = a_1 \cdot \dfrac{1-r^n}{1-r} \\S_{50} = \sqrt3 \cdot \left(\dfrac{1-(\sqrt3)^{50}}{1-\sqrt3}\right) \\S_{50} \approx 2004706374277$