Question

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

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

Answer 1

Answer: (A)

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
$> 100^{50}$
$\Rightarrow \left(101\right)^{50} > \left(100\right)^{50} + \left(99\right)^{50}$