evaluate: \n\\(\\sum_{n=1}^{7} 2\\left(\\frac{4}{3}\\right)^{n-1}\\)\n\\(s = ?\\)\nremember: for a geometric…

Explanation:

Step1: Identify geometric series parameters

For the series $\sum_{n = 1}^{7} 2\left(\frac{4}{3}\right)^{n - 1}$, the first term $a = 2$, common ratio $r=\frac{4}{3}$, and number of terms $n = 7$.

Step2: Apply geometric series formula

Use the formula $S=\frac{a(1 - r^{n})}{1 - r}$. Substitute $a = 2$, $r=\frac{4}{3}$, and $n = 7$: $$ \begin{align*} S&=\frac{2\left(1-\left(\frac{4}{3}\right)^{7}\right)}{1-\frac{4}{3}}\ &=\frac{2\left(1-\frac{16384}{2187}\right)}{-\frac{1}{3}}\ &=\frac{2\left(\frac{2187 - 16384}{2187}\right)}{-\frac{1}{3}}\ &=\frac{2\left(\frac{-14197}{2187}\right)}{-\frac{1}{3}}\ &=\frac{\frac{-28394}{2187}}{-\frac{1}{3}}\ &=\frac{28394}{2187}\times3\ &=\frac{28394}{729}\ &\approx38.95 \end{align*} $$ Wait, let's recalculate $\left(\frac{4}{3}\right)^7$: $4^7 = 16384$, $3^7=2187$, correct. Then $1-\frac{16384}{2187}=\frac{2187 - 16384}{2187}=\frac{-14197}{2187}$. Then numerator: $2\times\frac{-14197}{2187}=\frac{-28394}{2187}$. Denominator: $1 - \frac{4}{3}=-\frac{1}{3}$. So dividing by $-\frac{1}{3}$ is multiplying by $-3$? Wait, no: $\frac{\frac{-28394}{2187}}{-\frac{1}{3}}=\frac{-28394}{2187}\times(-3)=\frac{28394\times3}{2187}=\frac{85182}{2187}=\frac{28394}{729}\approx38.95$. Wait, but let's check the formula again. The geometric series formula for sum of first $n$ terms when $r\neq1$ is $S_n=\frac{a(r^n - 1)}{r - 1}$ (which is equivalent to $\frac{a(1 - r^n)}{1 - r}$). Let's use $S_n=\frac{a(r^n - 1)}{r - 1}$ to avoid negative signs. So $a = 2$, $r=\frac{4}{3}$, $n = 7$: $$ \begin{align*} S&=\frac{2\left(\left(\frac{4}{3}\right)^7 - 1\right)}{\frac{4}{3}-1}\ &=\frac{2\left(\frac{16384}{2187}-1\right)}{\frac{1}{3}}\ &=\frac{2\left(\frac{16384 - 2187}{2187}\right)}{\frac{1}{3}}\ &=\frac{2\left(\frac{14197}{2187}\right)}{\frac{1}{3}}\ &=\frac{28394}{2187}\times3\ &=\frac{28394}{729}\approx38.95 \end{align*} $$ Wait, but maybe I made a sign error. Let's re-express the formula: $S_n=\frac{a(1 - r^n)}{1 - r}$ when $r\lt1$, but here $r=\frac{4}{3}\gt1$, so the formula can also be written as $S_n=\frac{a(r^n - 1)}{r - 1}$, which is the same as $\frac{a(1 - r^n)}{1 - r}$ because $1 - r=-(r - 1)$, so $\frac{a(1 - r^n)}{1 - r}=\frac{a(r^n - 1)}{r - 1}$. So the calculation is correct. Let's compute $\frac{28394}{729}$: 729×38 = 27702, 28394 - 27702 = 692, so 38 + 692/729 ≈38 + 0.95 = 38.95. But let's check with another approach. Let's compute each term:

n=1: 2×(4/3)^0=2×1=2

n=2: 2×(4/3)^1=8/3≈2.6667

n=3: 2×(16/9)=32/9≈3.5556

n=4: 2×(64/27)=128/27≈4.7407

n=5: 2×(256/81)=512/81≈6.3210

n=6: 2×(1024/243)=2048/243≈8.4280

n=7: 2×(4096/729)=8192/729≈11.2373

Now sum them up: 2 + 8/3 = 14/3 ≈4.6667; +32/9 = 14/3 + 32/9 = 42/9 + 32/9 = 74/9 ≈8.2222; +128/27 = 74/9 + 128/27 = 222/27 + 128/27 = 350/27 ≈12.9630; +512/81 = 350/27 + 512/81 = 1050/81 + 512/81 = 1562/81 ≈19.2840; +2048/243 = 1562/81 + 2048/243 = 4686/243 + 2048/243 = 6734/243 ≈27.7119; +8192/729 = 6734/243 + 8192/729 = 20202/729 + 8192/729 = 28394/729 ≈38.95, which matches.

Answer:

$\frac{28394}{729}$ (or approximately 38.95)