Question 6

What is the value of $$S = \frac{1}{1 \times 3 \times 5} + \frac{1}{1 \times 4} + \frac{1}{3 \times 5 \times 7} + \frac{1}{4 \times 7} + \frac{1}{5 \times 7 \times 9 } + \frac{1}{7 \times 10}+.....$$upto 20 terms, then what is the value of S?

Solution

This series consists of two different series one having 2 numbers as product in denominator and the other has three numbers as product in the denominator
lat terms in each series is $$ \frac{1}{19 \times 21 \times 23}$$ and $$\frac{1}{28 \times 31}$$
$$\frac{1}{1 \times 3 \times 5} + \frac{1}{1 \times 4} + \frac{1}{3 \times 5 \times 7} + \frac{1}{4 \times 7} + \frac{1}{5 \times 7 \times 9 } + \frac{1}{7 \times 10}+.....$$upto 20 terms

=$$\frac{1}{1 \times 3 \times 5} +\frac{1}{3 \times 5 \times 7}+...\frac{1}{1 \times 4} +\frac{1}{4 \times 7}+..$$

=$$\frac{1}{4}( \frac{1}{1 \times 3 }- \frac{1}{3 \times 5}+ \frac{1}{5 \times 7}+\frac{1}{7 \times 9}...-\frac{1}{21 \times 23 }+\frac{1}{1 \times 3} (\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}....-\frac{1}{31})$$

=$$\frac{1}{4}( \frac{1}{1 \times 3 }-\frac{1}{21 \times 23})+\frac{1}{1 \times 3} (\frac{1}{1}-\frac{1}{31})$$

=$$\frac{40}{483}+\frac{10}{31}$$

=$$\frac{6070}{14973}$$


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