Question

Larger of $99^{50}+100^{50}$ and $101^{50}$ is

• A. $101^{50}$
• B. $99^{50}+100^{50}$
• C. both are equal
• D. none of these

Answer 1

Answer: (A)

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