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

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

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

and

b=101^{50}

, then

View answer and explanation

Correct option: A

a < b

Option A (

a < b

) is correct because

101^{50}-99^{50}

can be expanded using the binomial theorem, showing that

b

is greater than

a

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\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

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(1-x)^{201}\left(1+x+x^2\right)^{200}

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n