Posted by Lauren on Sunday, September 12, 2010 at 6:42pm.
the 1 is over k(k+1), and the A is over k, and the B is over k+1
common 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)]
[(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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