If a=∑∞n=0xn,b=∑∞n=0yn,c=∑∞n=0(xy)n where |x|,|y|<1 ; then
A
abc=a+b+c
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B
ab+bc=ac+b
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C
ac+bc=ab+c
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D
ab+ac=bc+a
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Solution
The correct option is Cac+bc=ab+c Clearly every summation is infinite series, a=11−x,b=11−y and c=11−xy or x=1−1a,y=1−1b Simplifying above equation we get, ac+bc=ab+c