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Arithmetic and geometric sequences
Series and Summation Notation

To add the terms of a sequence, we replace each comma between the terms with a plus sign, forming what is called a series. Because each sequence is infinite, the number of terms in the series associated with it is infinite also. Two examples of infinite series are:

12+22+32+42+...+n2+...

1+3+5+7+...+(2n+1)+...

There is a shorthand method of indicating the sum of the first n terms, or the nth partial sum of a sequence. This method, called summation notation, involves the symbol $\Sigma$ which is capital sigma in the Greek alphabet.

The expression $\displaystyle\sum_{n=1}^{5}\;n^2\;$ designates the sum of the five terms obtained if we successively substitute the natural numbers 1, 2, 3, 4 and 5 for n in the expression n2. Hence,

$\displaystyle\sum_{n=1}^{5}\;n^2\;=\;1^2+2^2+3^2+4^2+5^2\;=\;1+4+9+16+25\;=\;55\;$

Evaluate $\displaystyle\sum_{i=2}^{5}\;i^2\;$ and $\displaystyle\sum_{j=2}^{4}\;3j+4x\;$

$\displaystyle\sum_{i=2}^{5}\;i^2\;=\;2^2+3^2+4^2+5^2\;=\;4+9+16+25\;=\;54\;$

$\displaystyle\sum_{j=2}^{4}\;3j+4x\;=\;3\cdot2+4x+3\cdot3+4x+3\cdot4+4x=27+12x$

Given an=3n+1, find s4

s4 = a1+a2+a3+a4 =(3·1+1) + (3·2+1) + (3·3+1) + (3·4+1) = 4+7+10+13 = 34

Write the following in summation notation:

i) 8+27+64+125+216

ii) 4 + $\frac{9}{2}$ + $\frac{16}{3}$ + $\frac{25}{4}$ + $\frac{36}{5}$ + $\frac{49}{6}$

iii) 3 + $\frac{3}{4}$ + $\frac{1}{3}$ + $\frac{3}{16}$ + $\frac{1}{12}$ + $\frac{3}{49}$

Solution:

 a) 8+27+64+125+216 = 23+33+43+53+63= $\displaystyle\sum_{j=2}^{6}\;j^3\;=\;\displaystyle\sum_{i=1}^{5}\;(i+1)^3$ b) 4 + $\frac{9}{2}$ + $\frac{16}{3}$ + $\frac{25}{4}$ + $\frac{36}{5}$ + $\frac{49}{6}$ = $\frac{2^2}{1}$ + $\frac{3^2}{2}$ + $\frac{4^2}{3}$ + $\frac{5^2}{4}$ + $\frac{6^2}{5}$ + $\frac{7^2}{6}$ = $\displaystyle\sum_{i=1}^{6}\frac{(i+1)^2}{i}\;$ c) 3 + $\frac{3}{4}$ + $\frac{1}{3}$ + $\frac{3}{16}$ + $\frac{3}{25}$ + $\frac{1}{12}$ + $\frac{3}{49}$ = 3 + $\frac{3}{4}$ + $\frac{3}{9}$ + $\frac{3}{16}$ +$\frac{3}{25}$ + $\frac{3}{36}$ + $\frac{3}{49}$ = $\frac{3}{1^2}$ + $\frac{3}{2^2}$ + $\frac{3}{3^2}$ + $\frac{3}{4^2}$ + $\frac{3}{5^2}$ + $\frac{3}{6^2}$ + $\frac{3}{7^2}$ = $\displaystyle\sum_{i=1}^{7}\frac{3}{i^2}\;$