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If a =9950+10...

Question

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

A

a < b

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B

a = b

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C

a > b

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D

a - b = 100^{49}

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Solution

The correct option is A $a < b$
Using binomial theorem,
$101^{50} - 99^{50} = (100 + 1)^{50} - (100 - 1)^{50}$

$= (^{50} C_{0} 100^{50} +^{50} C_{1} 100^{49} +^{50} C_{2} 100^{48} + . . . . .) - (^{50} C_{0} 100^{50} -^{50} C_{1} 100^{49} +^{50} C_{2} 100^{48} - . . . . .)$
$= 2 (^{50} C_{1} 100^{49} +^{50} C_{3} 100^{47} +^{50} C_{5} 100^{45} + . . . . .)$
$= 2 (50 \cdot 100^{49} +^{50} C_{3} 100^{47} +^{50} C_{5} 100^{45} + . . . . .)$
$= 100 \cdot 100^{49} + 2 (^{50} C_{3} 100^{47} +^{50} C_{5} 100^{45} + . . . . .)$

$101^{50} - 99^{50} = 100^{50} + 2 (^{50} C_{3} 100^{47} +^{50} C_{5} 100^{45} + . . . . .) 101^{50} - 99^{50} > 100^{50}$

$101^{50} - 99^{50} > 100^{50} 101^{50} > 99^{50} + 100^{50} ∴ a < b$

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