Evaluate:
$x (x + y) + x^{2} (x^{2} + y^{2}) + x^{3} (x^{3} + y^{3}) +$ .......to n terms, where $x$ & $y$ both less than $1$

x^{2} \frac{(1 - x^{2 n})}{1 - x^{2}} + x y \frac{(1 - x^{n} y^{n})}{1 - x y}

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x^{2} \frac{(1 + x^{2 n})}{1 - x^{2}} + x y \frac{(1 - x^{n} y^{n})}{1 - x y}

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x^{2} \frac{(1 + x^{2 n})}{1 + x^{2}} + x y \frac{(1 + x^{n} y^{n})}{1 + x y}

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x^{2} \frac{(1 + x^{2 n})}{1 + x^{2}} + x y \frac{(1 - x^{n} y^{n})}{1 - x y}

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Solution

The correct option is A $x^{2} \frac{(1 - x^{2 n})}{1 - x^{2}} + x y \frac{(1 - x^{n} y^{n})}{1 - x y}$
$x (x + y) + x^{2} (x^{2} + y^{2}) + x^{3} (x^{3} + y^{3}) +$ .......to n terms, where $x$ & $y$ both less than $1$
Given, $(x^{2} + x^{4} + x^{6} + . . . .)$ n terms $+ (x y + x^{2} y^{2} + . . .)$ n terms
$= x^{2} \frac{(1 - x^{2 n})}{1 - x^{2}} + x y \frac{(1 - x^{n} y^{n})}{1 - x y}$
Hence, option A

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Similar questions

\frac{x (x + y) + x^{2} (x^{2} + y^{2}) + x^{3} (x^{3} + y^{3}) + . . . . t o n t e r m s}{x^{2} \frac{(1 - x^{2 n})}{1 - x^{2}} + x y \frac{(1 - x^{n} y^{n})}{1 - x y}}

(1 - x^{2}) (1 - y) d x = x y (1 + y) d y

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})

\frac{d y}{d x}

x = 0

1

y = \frac{1 - x^{2^{n + 1}}}{1 - x}

x y + y z + z x = 1

\frac{x}{1 - x^{2}} + \frac{y}{1 - y^{2}} + \frac{z}{1 - z^{2}} = \frac{4 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})}

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2 n}), then {(\frac{d y}{d x})}_{x = 0} =$

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