Evaluate 20(0.5)^(n-1) when n= 3 to 12. Use sigma notation .
what you have is a geometric series, with
a = 20
r = 1/2
S2 = 20(1-r^2)/(1-r)
S12 = 20(1-r^12)/1-r
sigma(3,12) = S12-S2
= 20[(1-r^12)/(1-r) - (1-r^2)/(1-r)]
= 20/(1-r) ((1-r^12)-(1-r^2))
= 5115/512
To evaluate the expression using sigma notation, let's define the expression as a function of n, and then use the sigma notation to calculate the summation.
Step 1: Define the function
Let's define the function f(n) as f(n) = 20(0.5)^(n-1)
Step 2: Write the sigma notation
The sigma notation represents the sum of a function over a range of values. In this case, we want to find the sum of f(n) for values of n ranging from 3 to 12.
∑[n=3 to 12] f(n)
Step 3: Evaluate the expression
To evaluate the summation, we substitute each value of n from 3 to 12 into the function f(n) and add up the values.
∑[n=3 to 12] 20(0.5)^(n-1)
= 20(0.5)^(3-1) + 20(0.5)^(4-1) + ... + 20(0.5)^(12-1)
= 20(0.5)^2 + 20(0.5)^3 + ... + 20(0.5)^11
To find the value of this summation, we simply calculate each term of the expression and add them up.
f(3) = 20(0.5)^2 = 20(0.25) = 5
f(4) = 20(0.5)^3 = 20(0.125) = 2.5
...
f(12) = 20(0.5)^11 = 20(0.00048828125) = 0.009765625
Now, let's add up all the terms:
5 + 2.5 + ... + 0.009765625
This summation can be calculated using a calculator or with mathematical software.