Science:Math Exam Resources/Courses/MATH101/April 2013/Question 02 (b)

MATH101 April 2013

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[hide] Question 02 (b)
Short-Answer Questions. Question 1-3 are short-answer questions. Put your answer in the box provided. Simplify your answer as much as possible. Full Marks will be awarded for a correct answer placed in the box. Show your work, for part marks. Find the sum of the series $\sum _{n=2}^{\infty }{\frac {3\cdot 4^{n+1}}{8\cdot 5^{n}}}$ Remember to simplify your answer completely.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Because we have to find the value of the series, we'll need a theorem which does more than determine convergence.

[show] Hint 2
What is the sum of a geometric series?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

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[show] Solution 1
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This is a geometric series. The first term is given by $\displaystyle n=2$ : $a={\frac {3\cdot 4^{3}}{8\cdot 5^{2}}}$ The common ratio is given by the quotient of successive terms. For example, using the first two terms gives: $r={\frac {\frac {3\cdot 4^{4}}{8\cdot 5^{3}}}{\frac {3\cdot 4^{3}}{8\cdot 5^{2}}}}={\frac {3\cdot 4^{4}}{8\cdot 5^{3}}}\cdot {\frac {8\cdot 5^{2}}{3\cdot 4^{3}}}={\frac {4}{5}}$ Then using the formula $a+ar+ar^{2}+\cdots ={\frac {a}{1-r}}$ we get $\sum _{n=2}^{\infty }{\frac {3\cdot 4^{n+1}}{8\cdot 5^{n}}}={\frac {\frac {3\cdot 4^{3}}{8\cdot 5^{2}}}{1-{\frac {4}{5}}}}={\frac {24}{5}}$

[show] Solution 2
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Simplifying the series helps to see the similarity to the geometric series: $\sum _{n=2}^{\infty }{\frac {3\cdot 4^{n+1}}{8\cdot 5^{n}}}={\frac {3\cdot 4}{8}}\sum _{n=2}^{\infty }\left({\frac {4}{5}}\right)^{n}$ Unlike the series above, the geometric series starts at $n=0$ , so we next rewrite our expression as ${\begin{aligned}{\frac {3\cdot 4}{8}}\sum _{n=2}^{\infty }\left({\frac {4}{5}}\right)^{n}&={\frac {3}{2}}\left(\sum _{n=0}^{\infty }\left({\frac {4}{5}}\right)^{n}-\sum _{n=0}^{1}\left({\frac {4}{5}}\right)^{n}\right)\\&={\frac {3}{2}}\left(\sum _{n=0}^{\infty }\left({\frac {4}{5}}\right)^{n}-\left(1+{\frac {4}{5}}\right)\right)\end{aligned}}$ Using the geometric series $\sum _{n=0}^{\infty }r^{n}={\frac {1}{1-r}}$ with $r=4/5$ we obtain ${\frac {3}{2}}\left(\sum _{n=0}^{\infty }\left({\frac {4}{5}}\right)^{n}-\left(1+{\frac {4}{5}}\right)\right)={\frac {3}{2}}\left({\frac {1}{1-4/5}}-\left(1+{\frac {4}{5}}\right)\right)$ Simplifying the expression above we arrive at the final answer $\sum _{n=2}^{\infty }{\frac {3\cdot 4^{n+1}}{8\cdot 5^{n}}}={\frac {24}{5}}$