Question

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

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

Answer 1

Answer: (A)

Since, $(101{)}^{50}=(100+1{)}^{50}$
$=100^{50}+{}^{50}{C}_{1}100^{49}+{}^{50}{C}_{2}100^{48}+\dots +1\dots \left(i\right)$
and ${\left(99\right)}^{50}={\left(100-1\right)}^{50}$
$=100^{50}-{}^{50}{C}_{1}100^{49}+{}^{50}{C}_{2}100^{48}-...+1\dots \left(ii\right)$
On subtracting $\left(ii\right)$ from $\left(i\right)$, we get
${\left(101\right)}^{50}-{\left(99\right)}^{50}=2\left\{{}^{50}{C}_{1}100^{49}+{}^{50}{C}_{3}100^{47}+\dots \right\}$
$=2\times {}^{50}{C}_{1}100^{49}$$+\left(2\times {}^{50}{C}_3\times 100^{47}+...\right)$
$=100\times 100^{49}+a$ positive number
$>100^{50}$
$\Rightarrow {\left(101\right)}^{50}>{\left(100\right)}^{50}+{\left(99\right)}^{50}$