Question

If $a={99}^{50}+{100}^{50}$ and $b={101}^{50}$, then :

• A. $a<b$
• B. $a=b$
• C. $a>b$
• D. $a-b={100}^{49}$

Answer 1

The correct option is A $a<b$

$a={99}^{50}+{100}^{50}$

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

$b{=}^{100}{C}_0{100}^{50}{+}^{100}{C}_1{100}^{49}+...{+}^{100}{C}_{100}{100}^0=2\times {100}^{50}{+}^{100}{C}_2{100}^{48}+....{+}^{100}{C}_{100}{100}^0$

$b={100}^{50}+{100}^{50}{+}^{100}{C}_2{100}^{48}+....{+}^{100}{C}_{100}{100}^0$

$b=a-{99}^{50}+{100}^{50}{+}^{100}{C}_2{100}^{48}+....{+}^{100}{C}_{100}{100}^0$

$\because ({100}^{50}-{99}^{50})>1$ and ${}^{100}{C}_2{100}^{48}+....{+}^{100}{C}_{100}{100}^0>1$

Hence, $b>a$