Science:Math Exam Resources/Courses/MATH105/April 2012/Question 01 (b)

MATH105 April 2012

• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q8 (e) • Q8 (f) • Q8 (g) • Q8 (h) • Q8 (i) •

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[hide] Question 01 (b)
You are given the formula for the sum of a geometric series, namely: $1+r+r^{2}+\cdots ={\frac {1}{1-r}},\qquad \|r\|<1.$ Use this fact to evaluate the series in part (a).

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Note that $1+r+r^{2}+\dots =\sum _{k=0}^{\infty }r^{k}$ Try to find a suitable value for r in our case.

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[show] Solution
The trick here is to recognize the series given as a geometric series. The question provides the formula $1+r+r^{2}+...=\sum _{k=0}^{\infty }r^{k}={\frac {1}{1-r}}$ for $\|r\|<1.$ Essentially, would like to evaluate the series by finding a suitable r so that this formula can be applied. Let's first re-express $\sum _{k=0}^{\infty }(-1)^{k}2^{k+1}x^{k}=\sum _{k=0}^{\infty }2(-1)^{k}2^{k}x^{k}$ as $2\sum _{k=0}^{\infty }(-1)^{k}2^{k}x^{k}$ so that all terms in the summation have an exponent of k and the indexing starts at 0. Now we note that $2\sum _{k=0}^{\infty }(-1)^{k}2^{k}x^{k}=2\sum _{k=0}^{\infty }[(-1)(2)(x)]^{k}=2\sum _{k=0}^{\infty }(-2x)^{k}$ . The formula provided now works with $r=-2x$ . The series sums to $2\times {\frac {1}{1-(-2x)}}={\frac {2}{1+2x}}.$