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August 20, 2014

August 20, 2014

Posted by **Lauren** on Sunday, September 12, 2010 at 6:42pm.

1 A B

----- = --- + ---

k(k+1) k k+1

- Pre Calc -
**Lauren**, Sunday, September 12, 2010 at 6:43pmthe 1 is over k(k+1), and the A is over k, and the B is over k+1

- Pre Calc -
**Anonymous**, Sunday, September 12, 2010 at 9:45pmcommon denominator: k(k+1)

1/[(k)(k+1)]=A(k+1)/[(k)(k+1)] + B(k)/[(k)(k+1)]

1/[(k)(k+1)]= [ A(k+1)+ B(k)]/ [(k)(k+1)]

cross multiply:

[(k)(k+1)]= [ A(k+1)+ B(k)][(k)(k+1)]

cancel like terms:

1= [ A(k+1)+ B(k)]

solve for k!

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