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  4. If x = 9950 + 10050 and y = (101)50 then

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Q. If $x = 99^{50} + 100^{50}$ and $y = \left(101\right)^{50}$ then

Binomial Theorem

Solution:

$\left(101\right)^{50}- \left(99\right)^{50} = \left(100 + 1\right)^{50}- \left(100-1\right) ^{50}$
$= 2\left[^{50}C_{1}\left(100\right)^{49}+^{50}C_{3}\left(100\right)^{47}+...... + ^{50}C_{49} \left(100\right)\right]$
$> 2. ^{50}C_{1} . \left(100\right)^{49} = 2 \times50\left(100\right)^{49} = \left(100\right)^{50}$
$\Rightarrow \left(101\right)^{50} > \left(99\right)^{50} + \left(100\right)^{50} \Rightarrow y > x \Rightarrow x < y .$