Find: (2/(35)) + (2/(63)) + (2/(99)) + (2/(143)) = ?


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Question Number 201430 by hardmath last updated on 06/Dec/23

$Find :$ $\frac{2}{35} + \frac{2}{63} + \frac{2}{99} + \frac{2}{143} = ?$
	Answered by Rasheed.Sindhi last updated on 06/Dec/23
	$2((1/(6^2 −1))+(1/(8^2 −1))+(1/(10^2 −1))+(1/(12^2 −1))) 2((1/((2(1)+4)^2 −1))+(1/((2(2)+4)^2 −1))+(1/((2(3)+4)^2 −1))+(1/((2(4)+4)^2 −1))) General term=(1/((2n+4)^2 −1)) =(1/((2n+4−1)(2n+4+1)))=(1/((2n+3)(2n+5))) Partial fraction: (a/(2n+3))+(b/(2n+5))=(1/((2n+3)(2n+5))) a(2n+5)+b(2n+3)=1 2an+2nb+5a+3b=1 (2a+2b)n+(5a+3b)=1 2a+2b=0 ∧ 5a+3b=1 b=−a ∧ 5a−3a=1⇒a=(1/2)⇒b=−(1/2) (1/((2n+3)(2n+5)))=((1/2)/(2n+3))+((−1/2)/(2n+5))=(1/2)((1/(2n+3))−(1/(2n+5))) =2Σ_(n=1) ^4 {(1/2)((1/(2n+3))−(1/(2n+5)))} =Σ_(n=1) ^4 ((1/(2n+3))−(1/(2n+5))) =Σ_(n=1) ^4 ((1/(2n+3)))−Σ_(n=1) ^4 ((1/(2n+5))) =((1/5)−(1/7))+((1/7)−(1/9))+((1/9)−(1/(11)))+((1/(11))−(1/(13))) =(1/5)−(1/(13))=(8/(65))$
	$2 (\frac{1}{6^{2} - 1} + \frac{1}{8^{2} - 1} + \frac{1}{10^{2} - 1} + \frac{1}{12^{2} - 1})$ $2 (\frac{1}{{(2 (1) + 4)}^{2} - 1} + \frac{1}{{(2 (2) + 4)}^{2} - 1} + \frac{1}{{(2 (3) + 4)}^{2} - 1} + \frac{1}{{(2 (4) + 4)}^{2} - 1})$ $G e n e r a l t e r m = \frac{1}{{(2 n + 4)}^{2} - 1}$ $= \frac{1}{(2 n + 4 - 1) (2 n + 4 + 1)} = \frac{1}{(2 n + 3) (2 n + 5)}$ $P a r t i a l f r a c t i o n :$ $\frac{a}{2 n + 3} + \frac{b}{2 n + 5} = \frac{1}{(2 n + 3) (2 n + 5)}$ $a (2 n + 5) + b (2 n + 3) = 1$ $2 a n + 2 n b + 5 a + 3 b = 1$ $(2 a + 2 b) n + (5 a + 3 b) = 1$ $2 a + 2 b = 0 \land 5 a + 3 b = 1$ $b = - a \land 5 a - 3 a = 1 \Rightarrow a = \frac{1}{2} \Rightarrow b = - \frac{1}{2}$ $\frac{1}{(2 n + 3) (2 n + 5)} = \frac{1 / 2}{2 n + 3} + \frac{- 1 / 2}{2 n + 5} = \frac{1}{2} (\frac{1}{2 n + 3} - \frac{1}{2 n + 5})$ $= 2 \underset{n = 1}{\sum^{4}} {\frac{1}{2} (\frac{1}{2 n + 3} - \frac{1}{2 n + 5})}$ $= \underset{n = 1}{\sum^{4}} (\frac{1}{2 n + 3} - \frac{1}{2 n + 5})$ $= \underset{n = 1}{\sum^{4}} (\frac{1}{2 n + 3}) - \underset{n = 1}{\sum^{4}} (\frac{1}{2 n + 5})$ $= (\frac{1}{5} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{9}) + (\frac{1}{9} - \frac{1}{11}) + (\frac{1}{11} - \frac{1}{13})$ $= \frac{1}{5} - \frac{1}{13} = \frac{8}{65}$
	Commented by hardmath last updated on 06/Dec/23

	$thank you professor cool$
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