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)

${101}^{50}$

We have ${101}^{50}={(100+1)}^{50}$

$={100}^{50}+50\times {100}^{49}+\frac{50\times 49}{1\times 2}\times {100}^{48}+...$

and ${99}^{50}={(100-1)}^{50}$

$={100}^{50}-50\times {100}^{49}+\frac{50\times 49}{1\times 2}\times {100}^{48}-...$

Subtracting, we get

${101}^{50}-{99}^{50}=2[50\times {100}^{49}+\frac{50\times 49\times 48}{1\times 2\times 3}\times {100}^{47}+...]$

$={100}^{50}+2\times \frac{50\times 49\times 48}{1\times 2\times 3}\times {100}^{47}+....>{100}^{50}$

Hence, ${101}^{50}>{99}^{50}+{100}^{50}$