What is the value of the sum \(2^{-1} + 2^{-2} + 2^{-3} + \cdots + 2^{-9} + 2^{-10}? \)Give your answer in simplest fraction.
+9673 I think we don't have to use any formulae.
Just observe.
\(2^{-1}+2^{-2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
\(2^{-1}+2^{-2}+2^{-3}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}\)
You can observe that this sum of finite geometric series = 1 - last term.
Therefore \(2^{-1}+2^{-2}+......+2^{-10}=1-2^{-10}=1-\frac{1}{1024}=\frac{1023}{1024}\)
LambLamb's answer is a good explanation of this phenomenon.
+495 What you have there is a finite geometric series.
The difference between the infinite sum of this geometric series and each consecutive partial sum has a ratio of 0.5, so you would take the infinite sum and subtract \({\frac{1}{2}}^{10}\).
The infinite sum is equal to exactly 1, and (1/2)^10 is equal to 1/1024, so the answer is
1023/1024.
The infinite sum is calculated by taking an analogous infinite sum (0.5)^0 + (0.5)^1 + (0.5)^2 + (0.5)^3 ... using the formula 1/(1-r) which the ratio is 0.5, making the sum equal to 1/0.5 = 2, then subtracting the first term (1) because it is not in the original finite geometric series.
Then by looking at partial sums, 0.5, 0.75, 0.875, etc, you can see the ratio between the difference of 1 and each consecutive partial sum is 0.5, so you can conclude 1 - (0.5)^10 will be the right answer.
Certainly not the most mathematically pure approach, but it works.
+9673 I think we don't have to use any formulae.
Just observe.
\(2^{-1}+2^{-2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
\(2^{-1}+2^{-2}+2^{-3}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}\)
You can observe that this sum of finite geometric series = 1 - last term.
Therefore \(2^{-1}+2^{-2}+......+2^{-10}=1-2^{-10}=1-\frac{1}{1024}=\frac{1023}{1024}\)
LambLamb's answer is a good explanation of this phenomenon.
