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,
$101^{50}=(100+1{)}^{50}$
$=100^{50}+50\cdot 100^{49}+\frac{50\cdot 49}{1\cdot 2}\cdot 100^{48}+\dots$
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
$101^{50}-99^{50}=2\left(50\cdot 100^{49}+\frac{50\cdot 49\cdot 48}{1\cdot 2\cdot 3}\times 100^{47}+\dots \right)$
$=100^{50}+2\cdot \frac{50\cdot 49\cdot 48}{1\cdot 2\cdot 3}\cdot 100^{47}+\dots >100^{50}$
Hence, $101^{50}>99^{50}+100^{50}$.