Q. The larger of $99^{50} + 100^{50}$ and $101^{50}$ is
A) $99^{50} + 100^{50}$
B) Both are equal
C) $101^{50}$
D) None of these
Explanation
(c)
We have, $(101)^{50} = (100 + 1)^{50}$
$= (100)^{50} + \ ^{50}C_{1}(100)^{49} + \ ^{50}C_{2}(100)^{48} + \ldots\ldots.$(i)
$and\ (99)^{50} = (100 - 1)^{50}$
$\ \ \ = (100)^{50} - \ ^{50}C_{1}(100)^{49} + \ ^{50}C_{2}(100)^{48} + \ldots\ldots$(ii)
On subtracting Eq. (ii) from Eq. (i), we get
$(101)^{50} - (99)^{50} = 2\{\ ^{50}C_{1}(100)^{49} + \ ^{50}C_{3}(100)^{47} + \ldots\}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 \times \ ^{50}C_{1}(100)^{49} + \{ 2 \times \ ^{50}C_{3}(100)^{47} + \ldots\}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = (100\left( 100)^{49} + a\ positive\ number \right) > (100)^{50}$
$\therefore(101)^{50} > (100)^{50} + (99)^{50}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.