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Logarithmic Function

If a=9950+1...

Question

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

a < b

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a = b

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a > b

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a - b = 100^{49}

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Solution

The correct option is A $a < b$
$a = 99^{50} + 100^{50}$
$b = 101^{50} = (100 + 1)^{50}$
$b =^{100} C_{0} 100^{50} +^{100} C_{1} 100^{49} + . . . +^{100} C_{100} 100^{0} = 2 \times 100^{50} +^{100} C_{2} 100^{48} + . . . . +^{100} C_{100} 100^{0}$

$b = 100^{50} + 100^{50} +^{100} C_{2} 100^{48} + . . . . +^{100} C_{100} 100^{0}$
$b = a - 99^{50} + 100^{50} +^{100} C_{2} 100^{48} + . . . . +^{100} C_{100} 100^{0}$
$∵ (100^{50} - 99^{50}) > 1$ and $^{100} C_{2} 100^{48} + . . . . +^{100} C_{100} 100^{0} > 1$

Hence, $b > a$

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