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

MCQ

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

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

✓

Answer

(a) x < y
Explanation: 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
$\begin{array}{l}(101)^{50}-(99)^{50}=2\left[{ }^{50} C_1 \quad(100)^{49}+{ }^{50} C_3 \quad(100)^{47}+\ldots\right] \\ =2\left[50(100)^{49}+\frac{50 \times 49 \times 48}{3 \times 2 \times 1}(100)^{47}+\ldots\right] \\ =(100)^{50}+2\left(\frac{50 \times 49 \times 48}{3 \times 2 \times 1}(100)^{47}\right) \\ \Rightarrow(101)^{50}-(99)^{50}>(100)^{50}\end{array}$
$\begin{array}{l}\Rightarrow(101)^{50} > (100)^{50}+(99)^{50} \\
\Rightarrow y > x\end{array}$

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