Question

Given that |x|$<$1 and |y|$<$1, then find the sum of the series $x(x+y)+x^2(x^2+y^2)+x^3(x^3+y^3).....$

• A.
• B.
• C. 2xy
• D.

Answer 1

The correct option is B

Start out by assuming values for x and y

Say $x=\frac{1}{2}$ and $y=\frac{1}{2}$

Find the sum of the first two terms of the series $=\frac{1}{2}+\frac{1}{8}=\frac{5}{8}=0.63$ (Approx)

The sum of the series will be slightly more than 0.63 as the denominator will keep increasing and the value of each term will reduce.

Look at the answer options

Substitute values of $x=\frac{1}{2}$ and $y=\frac{1}{2}$

Option $a=\frac{x{y}^2}{(1-x{y}^2)}=\frac{1}{7}=0.16$

Option $b=\frac{\left(x^2\right)}{(1-x^2)}+\frac{\left(xy\right)}{(1-xy)}$

$\frac{\left(\frac{4}{3}\right)}{4}+\frac{\left(\frac{4}{3}\right)}{4}$

Answer = $\frac{2}{3}$ = 0.667 (Approx)

Option c = 2xy = $\frac{2}{4}$ =$\frac{1}{2}$ = 0.5

Option d = 2xy + $\frac{1}{x}$ = 0.5 + 2 = 2.5

Answer can never be (a), option (c) , and option (d) is too high. Answer = option (b)