Answer:
![(468)/(25)](https://img.qammunity.org/2023/formulas/mathematics/college/6ij6d94525smdiqg2bpagucvmguqwrqqco.png)
Explanation:
We are asked to find
![\displaystyle \sum _(n=0)^33\left((1)/(5)\right)^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/2412do8rrsbz6madr88h5kx72dscsxq0b1.png)
We are asked to find s₄ which is the sum of the first 4 terms of the series. Since n starts at 0, this would be
a₀ + a₁ + a₂ + a₃
We can compute the individual terms and add them up
![n = 0 \\\rightarrow \displaystyle a_0 = 3\left((1)/(5)\right)^(0-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(-1)\\\\\rightarrow \displaystyle 3\left(5}\right)^(1)\\ \rightarrow \displaystyle 15](https://img.qammunity.org/2023/formulas/mathematics/college/4gefg6i9q0xdsmtof5xxmsrdgvbelt2z18.png)
This would be the first term of the geometric sequence
n = 1
![\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(1-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(0)\\\\\rightarrow \displaystyle 3\cdot 1}\\ \rightarrow \displaystyle a_1 = 3](https://img.qammunity.org/2023/formulas/mathematics/college/llzd5uygh0xgqbu7nkru1viwwsy3olivtp.png)
n = 2
![\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(2-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(1)\\\\\rightarrow \displaystyle a_2 = (3)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/84xje37ixe28xvkeplw1nb9grvey2xajzo.png)
n = 3
![\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(3-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(2)\\\\\rightarrow \displaystyle a_3 = (3)/(25)](https://img.qammunity.org/2023/formulas/mathematics/college/f9lye7u51zkffj7zt1pjsmo91plrr7d2cv.png)
Adding these we get
15 + 3 + 3/5 + 3/25 =
![(468)/(25)](https://img.qammunity.org/2023/formulas/mathematics/college/6ij6d94525smdiqg2bpagucvmguqwrqqco.png)
Alternate route
In a geometric sequence the sum of the formula for the first n terms will be
![a_0 + a_1(1-r^(n))/(1-r)](https://img.qammunity.org/2023/formulas/mathematics/college/drubd8b9bw7e8d2jqu1avfopx04a9uu8kk.png)
We computed a₀ as 15 and a₁ as 3
So sum
![= 15 + 3\left((1-\left((1)/(5)\right)^3)/(1-(1)/(5))\right)\\](https://img.qammunity.org/2023/formulas/mathematics/college/7ush0a6rqfe7bzz51oc84zc7c0n5l25mz0.png)
If you evaluate this, it should work out to the same fraction:
![(468)/(25)](https://img.qammunity.org/2023/formulas/mathematics/college/6ij6d94525smdiqg2bpagucvmguqwrqqco.png)