evaluate the R(1(3))
Answers
They've given me the sum of the first four terms, S4, and the value of the common ratio r. Since there is a common ratio, I know this must be a geometric series. Plugging into the geometric-series-sum formula, I get:
\mathrm{S}_4 = a\left(\dfrac{1 - r^4}{1 - r}\right)S4=a(1−r1−r4)
\small{\dfrac{26}{27}} = a\left(\dfrac{1 - \left(\frac{1}{3}\right)^4}{1 - \frac{1}{3}}\right)2726=a(1−311−(31)4)
= a\left(\dfrac{\left(\frac{80}{81}\right)}{\left(\frac{2}{3}\right)}\right)=a((32)(8180))
= a\left(\small{\dfrac{80}{81}}\right)\left(\small{\dfrac{3}{2}}\right)=a(8180)(23)
= a\left(\small{\dfrac{40}{27}}\right)=a(2740)
Multiplying on both sides by \frac{27}{40}4027 to solve for the first term a = a1, I get:
\left(\small{\dfrac{26}{27}}\right)\left(\small{\dfrac{27}{40}}\right) = a = \small{\dfrac{13}{20}}(2726)(4027)=a=2013
Then, plugging into the formula for the n-th term of a geometric sequence, I get:
\color{purple}{\mathbf{ \mathit{a}_{\mathit{n}} = \small{\dfrac{13}{20}}\left(\small{\dfrac{1}{3}}\right)^{\mathit{n}-1} }}an=2013(31)n−1
Show, by use of a geometric series, that 0.3333... is equal to \mathbf{\color{green}{\frac{1}{3}}}31.
There's a trick to this. I first have to break the repeating decimal into separate terms; that is, "0.3333..." becomes:
0.3 + 0.03 + 0.003 + 0.0003 + ...
Splitting up the decimal form in this way highlights the repeating pattern of the non-terminating (that is, the never-ending) decimal explicitly: For each term, I have a decimal point, followed by a steadily-increasing number of zeroes, and then ending with a "3". This expanded-decimal form can be written in fractional form, and then converted into geometric-series form:
0.3333... = 0.3 + 0.03 + 0.003 + 0.0003 + ...0.3333...=0.3+0.03+0.003+0.0003+...
= \frac{3}{10} + \frac{3}{100} + \frac{3}{1,000} + \frac{3}{10,000} + ...=103+1003+1,0003+10,0003+...
= \frac{3}{10} + \frac{3}{10}\left(\frac{1}{10}\right) + \frac{3}{10}\left(\frac{1}{100}\right) + \frac{3}{10}\left(\frac{1}{1,000}\right) + ...=103+103(101)+103(1001)+103(1,0001)+...
= \frac{3}{10} + \frac{3}{10}\left(\frac{1}{10}\right) + \frac{3}{10}\left(\frac{1}{10}\right)^2 + \frac{3}{10}\left(\frac{1}{10}\right)^3 + ...=103+103(101)+103(101)2+103(101)3+...
= \displaystyle{ \sum_{i=1}^{\infty}\, \small{\dfrac{3}{10}}\left(\small{\dfrac{1}{10}}\right)^{n-1} }=i=1∑∞103(101)n−1
This proves that 0.333... is (or, at least, can be expressed as) an infinite geometric series with a = \frac{3}{10}a=10