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Question

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 C ac+bc=ab+c
Clearly every summation is infinite series,
a=11x,b=11y and c=11xy
or x=11a,y=11b
Simplifying above equation we get, ac+bc=ab+c

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