If x=99^ 50 +100^ 50 and y=(101)^ 50 then - Vidyadip

MCQ

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

✓
$x < y$
B
$x > y$
C
$x = y$
D
$x \geq y$

✓

Answer

Correct option: A.

$x < y$

Given $x=99^{50}+100^{50}$ and $y=(101)^{50}$
Now $y, =(101)^{50}=(100+1)^{50}={ }^{50} C_0(100)^{50}+{ }^{50} C_1(100)^{49}+{ }^{50} C_2(100)^{48}+\ldots .+{ }^{50} C_{50} \ldots .$
$(i)$ Also $(99)^{50}=(100-1)^{50}=={ }^{50} C_0(100)^{50}-{ }^{50} C_1(100)^{49}+{ }^{50} C_2(100)^{48}-\ldots .+{ }^{50} C_{50} \ldots$
$(ii)$ Now subtract equation (ii) from equation $(i),$
we get $(101)^{50}-(99)^{50}=2\left[{ }^{50} C_1 \quad(100)^{49}+{ }^{50} C_3 \quad(100)^{47}+\ldots\right]$
$=2\left[50(100)^{49}+\frac{50 \times 49 \times 48}{3 \times 2 \times 1}(100)^{47}+\ldots\right]$
$=(100)^{50}+2\left(\frac{50 \times 49 \times 48}{3 \times 2 \times 1}(100)^{47}\right)$
$\Rightarrow(101)^{50}-(99)^{50}>(100)^{50}$
$\Rightarrow(101)^{50} > (100)^{50}+(99)^{50}$
$\Rightarrow y > x$

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