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Question

Mathematics Question on Sequence and series

If x=1+a+a2+....................x = 1 + a + a^2 + .................... to infinity and y=1+b+b2+...................y = 1 + b + b^2 + ................... to infinity, where a, b are proper fractions, then 1+ab+a2b2+.....1 + ab + a^2b^2 + ..... to infinity is equal :

A

xyx+y1\frac{xy}{x+y-1}

B

xyxy1\frac{xy}{x-y-1}

C

xyxy+1\frac{xy}{x-y+1}

D

xyx+y+1\frac{xy}{x+y+1}

Answer

xyx+y1\frac{xy}{x+y-1}

Explanation

Solution

If a,ar,ar2,ar3.........a, ar, ar^2, ar^3 ......... are in G.P., then sum of infinite GP=a+ar+.....+=a1rGP=a+ar+.....+\infty=\frac{a}{1-r} where 'a' is the first term and 'r' is the common ratio of G.P. Given x=1+a+a2+.....x = 1+ a + a^2 + .....\infty This is a GP, with common ratio 'a'. x=11axax=1a=x1x\Rightarrow x=\frac{1}{1-a} \Rightarrow x-ax=1 \Rightarrow a=\frac{x-1}{x} Again, y=1+b+b2+......y = 1+ b + b^{2} + ......\infty This is also a G.P., with common ratio 'b'. y=11bb=y1y\Rightarrow y=\frac{1}{1-b} \Rightarrow b=\frac{y-1}{y} Now, consider 1+ab+a2b2+.....1+ab+a^{2}b^{2}+.....\infty which is again a GP with common ratio 'ab'. \therefore Sum 11ab=11x1x.y1y-\frac{1}{1-ab}=\frac{1}{1-\frac{x-1}{x}. \frac{y-1}{y}} =xyxyxy+x+y1=xyx+y1=\frac{xy}{xy-xy+x+y-1}=\frac{xy}{x+y-1}