First slide

Binomial theorem for positive integral Index

Question

Larger of 99⁵⁰ + 100⁵⁰ and 101⁵⁰ is

Moderate

A

101⁵⁰
B

99⁵⁰ + 100⁵⁰
C

both are equal
D

none of these

Solution

We have,
$\begin{aligned} 101^{50} = (100 + 1)^{50} \\ = 100^{50} + 50 \cdot 100^{49} + \frac{50 \cdot 49}{1 \cdot 2} \cdot 100^{48} + \dots \end{aligned}$
and $99^{50} = (100 - 1)^{50}$
$= 100^{50} - 50 \cdot 100^{49} + \frac{50 \cdot 49}{1 \cdot 2} \cdot 100^{48} - \dots$
Subtracting, we get
$\begin{aligned} 101^{50} - 99^{50} = 2 (50 \cdot 100^{49} + \frac{50 \cdot 49 \cdot 48}{1 \cdot 2 \cdot 3} \times 100^{47} + \dots) \\ = 100^{50} + 2 \cdot \frac{50 \cdot 49 \cdot 48}{1 \cdot 2 \cdot 3} \cdot 100^{47} + \dots > 100^{50} \end{aligned}$
Hence, $101^{50} > 99^{50} + 100^{50}$

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