Question

if 2/3=[x-1/y] + [x^2-1/y^2] + [x^3-1/y^3]+...... upto where xy =2 and |x| <1, calculate the value of x and y

Answer 1

Given Equation is:

23=(x−1y)+(x2−1y2)+...

Given conditions are:

xy=2and|x|<1

Hence we can say that |y|>2⟹1|y|<12

Now simplifying the given equation:

23=(x+x2+x3+...)−(1y+1y2+1y3+...)

Both of these RHS terms form an infinite GP, where |r|<1

So, sum of terms of an infinite GP with |r|<1 is given by Sn=a1−r

So, the given equation is simplified to:

23=x1−x−1y1−1y

⟹23=x1−x−1y−1

⟹23=x(y−1)−(1−x)(1−x)(y−1)

⟹23=xy−x−1+xy−1−xy+x

⟹23=2−1x+y−1−2

⟹23=1x+y−3

⟹2(x+y−3)=3⟹2x+2y−6=3

⟹2x+2y=9

But we know that y=2x⟹2y=4x

Hence the equation becomes:

2x+4x=9

⟹2x2+4=9x⟹2x2−9x+4=0

⟹2x2−x−8x+4=0⟹x(2x−1)−4(2x−1)

⟹(x−4)(2x−1)=0⟹x=4orx=12

So if x=4 the y=12 and if x=12 then y=4