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how to break 1/(2n+1)(2n+3)(2n+5) into partial fraction
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\[\frac{ 1 }{ (2n+1)(2n+3)(2n+5) }\]
Let.. \[\Large \frac{ 1}{(2n+1)(2n+3)(2n+5) } = \frac{A}{2n+1} + \frac{B}{2n+3} + \frac{C}{2n+5}\]
take LCM..solve for A B and C..substitute
ok, got it :)
glad :)
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can i solve it by multiplying by (2n+1)
@hartnn
what do u mean ? why only multiply by 2n+1 ?
|dw:1404666844270:dw|
now substituting by n by -1/2 so getting the value of A
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but that won't help you get A,B,C values
repeating the
process again
ok, you can do that
after taking LCM also, you'd put x=-1/2 to find A, its same thing, just different ways of doing.
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