Use the explicit formula to generate the first 4 terms of each geometric sequence. 1 g_(n)=2cdot 3^n-1 2. g_(n)=8240cdot 1.05^n-1 3. g_(n)=100cdot ((1)/(2))^n-1 4. g_(n)=(-2)cdot 4^n-1

1. For the geometric sequence $g_{n}=2\cdot 3^{n-1}$, we can generate the first 4 terms by substituting $n=1, 2, 3, 4$ into the formula:<br /> - $g_{1}=2\cdot 3^{1-1}=2\cdot 3^{0}=2\cdot 1=2$<br /> - $g_{2}=2\cdot 3^{2-1}=2\cdot 3^{1}=2\cdot 3=6$<br /> - $g_{3}=2\cdot 3^{3-1}=2\cdot 3^{2}=2\cdot 9=18$<br /> - $g_{4}=2\cdot 3^{4-1}=2\cdot 3^{3}=2\cdot 27=54$<br /><br /> Therefore, the first 4 terms of the geometric sequence $g_{n}=2\cdot 3^{n-1}$ are 2, 6, 18, and 54.<br /><br />2. For the geometric sequence $g_{n}=8240\cdot 1.05^{n-1}$, we can generate the first 4 terms by substituting $n=1, 2, 3, 4$ into the formula:<br /> - $g_{1}=8240\cdot 1.05^{1-1}=8240\cdot 1.05^{0}=8240\cdot 1=8240$<br /> - $g_{2}=8240\cdot 1.05^{2-1}=8240\cdot 1.05^{1}=8240\cdot 1.05=8642$<br /> - $g_{3}=8240\cdot 1.05^{3-1}=8240\cdot 1.05^{2}=8240\cdot 1.1025=9044.4$<br /> - $g_{4}=8240\cdot 1.05^{4-1}=8240\cdot 1.05^{3}=8240\cdot 1.157625=9525.1$<br /><br /> Therefore, the first 4 terms of the geometric sequence $g_{n}=8240\cdot 1.05^{n-1}$ are 8240, 8642, 9044.4, and 9525.1.<br /><br />3. For the geometric sequence $g_{n}=100\cdot (\frac {1}{2})^{n-1}$, we can generate the first 4 terms by substituting $n=1, 2, 3, 4$ into the formula:<br /> - $g_{1}=100\cdot (\frac {1}{2})^{1-1}=100\cdot (\frac {1}{2})^{0}=100\cdot 1=100$<br /> - $g_{2}=100\cdot (\frac {1}{2})^{2-1}=100\cdot (\frac {1}{2})^{1}=100\cdot \frac {1}{2}=50$<br /> - $g_{3}=100\cdot (\frac {1}{2})^{3-1}=100\cdot (\frac {1}{2})^{2}=100\cdot \frac {1}{4}=25$<br /> - $g_{4}=100\cdot (\frac {1}{2})^{4-1}=100\cdot (\frac {1}{2})^{3}=100\cdot \frac {1}{8}=12.5$<br /><br /> Therefore, the first 4 terms of the geometric sequence $g_{n}=100\cdot (\frac {1}{2})^{n-1}$ are 100, 50, 25, and 12.5.<br /><br />4. For the geometric sequence $g_{n}=(-2)\cdot 4^{n-1}$, we can generate the first 4 terms by substituting $n=1, 2, 3, 4$ into the formula:<br /> - $g_{1}=(-2)\cdot 4^{1-1}=(-2)\cdot 4^{0}=(-2)\cdot 1=(-2)$<br /> - $g_{2}=(-2)\cdot 4^{2-1}=(-2)\cdot 4^{1}=(-2)\cdot 4=(-8)$<br /> - $g_{3}=(-2)\cdot 4^{3-1}=(-2)\cdot 4^{2}=(-2)\cdot 16=(-32)$<br /> - $g_{4}=(-2)\cdot 4^{4-1}=(-2)\cdot 4^{3}=(-2)\cdot 64=(-128)$<br /><br /> Therefore, the first 4 terms of the geometric sequence $g_{n}=(-2)\cdot 4^{n-1}$