lf $x = (99)^{50} + (100)^{50}$ and $y = (101)^{50}$ then

x > y

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x < y

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x = y

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x y = 1

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Solution

The correct option is B $x < y$
$y - x$
$= (101)^{50} - (99)^{50} - (100)^{50}$

$= (100 + 1)^{50} - (100 - 1)^{50} - 100^{50}$

$= [(100 + 1)^{50} - (100 - 1)^{50}] - 100^{50}$

$= 2 [100^{49}^{50} C_{49} + 100^{47}^{50} C_{47} + . . . .100^{50} C_{1}] - 100^{50}$
$= 100^{49}^{50} C_{49} - 100^{50} + 100^{49}^{50} C_{49} + 2 [100^{47}^{50} C_{47} + 100^{45}^{50} C_{45} + . . .100^{50} C_{1}]$

$= 100^{49} [^{50} C_{49} - 100] + 100^{49}^{40} C_{49} + 2 [100^{47}^{50} C_{47} + 100^{45}^{50} C_{45} + . . .100^{50} C_{1}]$

$= 100^{49} (50 - 100) + 100^{49} (50) + 200 K$ where $k$ is a positive integer.

$= - {50.100}^{49} + 100^{49} (50) + 200 k$

$= 200 k$
Hence
$y - x > 0$
$y > x$

Suggest Corrections

Similar questions

x = {(99)}^{50} + {(100)}^{50}

y = {(101)}^{50}

(A) x > y

(B) x < y

y = e^{a {cos}^{- 1} x}

(1 - x^{2}) y^{''} - x y^{'} =

y = sin (7 {sin}^{- 1} x)

(1 - x^{2}) y^{''} - x y^{'} =

a = 99^{50} + 100^{50}

b = 101^{50}

99^{50} + 100^{50}

101^{50}

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