Question

Evaluate the sum of the following finite geometric series.

Answer 1

`\large\boxed{(156)/(125)\approx1.2}`

Explanation:

Method 1:

`\sum\limits_(n=1)^4\left((1)/(5)\right)^(n-1)\\\\for\ n=1\\\\\left((1)/(5)\right)^(1-1)=\left((1)/(5)\right)^0=1\\\\for\ n=2\\\\\left((1)/(5)\right)^(2-1)=\left((1)/(5)\right)^1=(1)/(5)\\\\for\ n=3\\\\\left((1)/(5)\right)^(3-1)=\left((1)/(5)\right)^2=(1)/(25)\\\\for\ n=4\\\\\left((1)/(5)\right)^(4-1)=\left((1)/(5)\right)^3=(1)/(125)`

`\sum\limits_(n=1)^4\left((1)/(5)\right)^(n-1)=1+(1)/(5)+(1)/(25)+(1)/(125)=(125)/(125)+(25)/(125)+(5)/(125)+(1)/(125)=(156)/(125)`

Method 2:

`\sum\limits_(n=1)^4\left((1)/(5)\right)^(n-1)\to a_n=\left((1)/(5)\right)^(n-1)\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot(1-r^n)/(1-r)\\\\r-\text{common ratio}\to r=(a_(n+1))/(a_n)\\\\a_(n+1)=\left((1)/(5)\right)^(n+1-1)=\left((1)/(5)\right)^n\\\\r=(\left((1)/(5)\right)^n)/(\left((1)/(5)\right)^(n-1))\qquad\text{use}\ (a^n)/(a^m)=a^(n-m)\\\\r=\left((1)/(5)\right)^(n-(n-1))=\left((1)/(5)\right)^(n-n+1)=\left((1)/(5)\right)^1=(1)/(5)`

`a_1=\left((1)/(5)\right)^(1-1)=\left((1)/(5)\right)^0=1`

`\text{Substitute}\ a_1=1,\ n=4,\ r=(1)/(5):\\\\S_4=1\cdot(1-\left((1)/(5)\right)^4)/(1-(1)/(5))=(1-(1)/(625))/((4)/(5))=(624)/(625)\cdot(5)/(4)=(156)/(125)`