Science:Math Exam Resources/Courses/MATH103/April 2017/Question 02 (a)

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MATH103 April 2017

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Question 02 (a)
Write the following in $\sum$ -notation: (i) $1+2+{\frac {9}{4}}+{\frac {16}{8}}+{\frac {25}{16}}+{\frac {36}{32}}=$ (ii) $1-2+16-2^{9}+2^{16}=$

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Hint
Find a pattern to the individual terms in each explicit summation.

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Solution
(i) In the explicit summation, we can find that the numerators are square numbers i.e., $n^{2}$ and the denominators are of the form $2^{n}$ . Based on this observation, we can rewrite the given summation using $\sum$ -notation as follows: $1+2+{\frac {9}{4}}+{\frac {16}{8}}+{\frac {25}{16}}+{\frac {36}{32}}={\frac {1}{2^{0}}}+{\frac {2^{2}}{2^{1}}}+{\frac {3^{2}}{2^{2}}}+{\frac {4^{2}}{2^{3}}}+{\frac {5^{2}}{2^{4}}}+{\frac {6^{2}}{2^{5}}}=\sum _{n=0}^{5}{\frac {(n+1)^{2}}{2^{n}}}$ This sum can also be written in the form $\sum _{n=1}^{6}{\frac {n^{2}}{2^{n-1}}}.$ (ii) First, we can observe that each term is a power of 2, so that we can write the given sum as $1-2+16-2^{9}+2^{16}=2^{0}-2^{1}+2^{4}-2^{9}+2^{16}$ Then, we can see that the exponents are square numbers, and hence $2^{0}-2^{1}+2^{4}-2^{9}+2^{16}=2^{0^{2}}-2^{1^{2}}+2^{2^{2}}-2^{3^{2}}+2^{4^{2}}.$ Therefore, the expression using $\sum$ -notation would be $1-2+16-2^{9}+2^{16}=2^{0^{2}}-2^{1^{2}}+2^{2^{2}}-2^{3^{2}}+2^{4^{2}}=\sum _{n=0}^{4}(-1)^{n}2^{n^{2}}$ . Answer: $\color {blue}(i)\sum _{n=0}^{5}{\frac {(n+1)^{2}}{2^{n}}},(ii)\sum _{n=0}^{4}(-1)^{n}2^{n^{2}}$