Question : What is the value of S $=\frac{1}{1×3×5}+\frac{1}{1×4}+\frac{1}{3×5×7}+\frac{1}{4×7}+\frac{1}{5×7×9}+\frac{1}{7×10}+....$ Up to 20 terms, then what is the value of S?
Option 1: $\frac{6179}{15275}$
Option 2: $\frac{6070}{14973}$
Option 3: $\frac{7191}{15174}$
Option 4: $\frac{5183}{16423}$
Correct Answer: $\frac{6070}{14973}$
Solution :
Given:
$S=\frac{1}{1×3×5}+\frac{1}{1×4}+\frac{1}{3×5×7}+\frac{1}{4×7}+\frac{1}{5×7×9}+\frac{1}{7×10}+....$ Up to 20 terms.
Rewriting it like,
$\frac{1}{1×3×5}+\frac{1}{3×5×7}+\frac{1}{5×7×9}+...+\frac{1}{1×4}+\frac{1}{4×7}+\frac{1}{7×10}+...$Up to 20 terms.
$=\frac{1}{4}[\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+...–\frac{1}{21×23}]+\frac{1}{3}[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...-\frac{1}{31}]$
It will cancel out most of the terms except the first and last term.
$=\frac{1}{4}[\frac{1}{1×3}-\frac{1}{21×23}]+\frac{1}{3}[1-\frac{1}{31}]$
$=\frac{40}{483}+\frac{10}{31}$
$=\frac{6070}{14973}$
Hence, the correct answer is $\frac{6070}{14973}$.
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