Question

$(100)^{50} + (99)^{50}$

• A. $< (101)^{50}$
• B. $= (101)^{50}$
• C. $> (101)^{50}$
• D. $> (101)^{51}$

Answer 1

Answer: (A)

$< (101)^{50}$

Answer 2

Since, $(101)^{50} = (100 + 1)^{50}$ $= 100^{50} + \,^{50}C_{1}\,100^{49} + \,^{50}C_{2}\,100^{48} + \ldots + 1\quad\ldots\left(i\right)$ and $\left(99\right)^{50} = \left(100-1\right)^{50}$ $= 100^{50 }- \,^{50}C_{1}\,100^{49} + \,^{50}C_{2}\,100^{48} -... +1 \quad\ldots\left(ii\right)$ On subtracting $\left(ii\right)$ from $\left(i\right)$, we get \left(101\right)^{50}-\left(99\right)^{50}= 2\left\\{^{50}C_{1}\,100^{49}+\,^{50}C_{3}\,100^{47}+\ldots\right\\} $= 2\times\,^{50}C_{1}\,100^{49}$ $+\left(2 \times \,^{50}C_{3} \times 100^{47}+...\right)$ $=100 \times100^{49}+a$ positive number $> 100^{50}$ $\Rightarrow \left(101\right)^{50} > \left(100\right)^{50} + \left(99\right)^{50}$