What is the value of S = frac{1}{1 times 3 ti | vedclass

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(B) The series consists of $20$ terms,which can be grouped into two separate series,each having $10$ terms.Let $S = S_1 + S_2$,where $S_1 = \sum_{n=1}

Vedclass · Accepted Answer

$\frac{6070}{14973}$

Vedclass · Answer

(B) The series consists of $20$ terms,which can be grouped into two separate series,each having $10$ terms.Let $S = S_1 + S_2$,where $S_1 = \sum_{n=1}^{10} \frac{1}{(2n-1)(2n+1)(2n+3)}$ and $S_2 = \sum_{n=1}^{10} \frac{1}{(3n-2)(3n+1)}$.For $S_1$,the general term is $a_n = \frac{1}{(2n-1)(2n+1)(2n+3)} = \frac{1}{4} [\frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)}]$.Summing $S_1$ from $n=1$ to $10$: $S_1 = \frac{1}{4} [(\frac{1}{1 	imes 3} - \frac{1}{3 	imes 5}) + (\frac{1}{3 	imes 5} - \frac{1}{5 	imes 7}) + \dots + (\frac{1}{19 	imes 21} - \frac{1}{21 	imes 23})] = \frac{1}{4} [\frac{1}{3} - \frac{1}{483}] = \frac{1}{4} [\frac{161-1}{483}] = \frac{160}{4 	imes 483} = \frac{40}{483}$.For $S_2$,the general term is $b_n = \frac{1}{(3n-2)(3n+1)} = \frac{1}{3} [\frac{1}{3n-2} - \frac{1}{3n+1}]$.Summing $S_2$ from $n=1$ to $10$: $S_2 = \frac{1}{3} [(1 - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{7}) + \dots + (\frac{1}{28} - \frac{1}{31})] = \frac{1}{3} [1 - \frac{1}{31}] = \frac{1}{3} [\frac{30}{31}] = \frac{10}{31}$.Total sum $S = S_1 + S_2 = \frac{40}{483} + \frac{10}{31} = \frac{40 	imes 31 + 10 	imes 483}{483 	imes 31} = \frac{1240 + 4830}{14973} = \frac{6070}{14973}$.

What is the value of $S = \frac{1}{1 \times 3 \times 5} + \frac{1}{1 \times 4} + \frac{1}{3 \times 5 \times 7} + \frac{1}{4 \times 7} + \frac{1}{5 \times 7 \times 9} + \frac{1}{7 \times 10} + \dots$ up to $20$ terms?

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