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Question

Mathematics Question on Binomial theorem

If 21st21^{st} and 22nd22^{nd} terms in the expansion of (1+x)44(1 + x)^{44} are equal, then xx is equal to

A

2122\frac {21}{22}

B

2324\frac {23}{24}

C

87\frac {8}{7}

D

78\frac {7}{8}

Answer

78\frac {7}{8}

Explanation

Solution

Given expansion is (1+x)44(1+x)^{44}.
According to the question, T21=T22T_{21}=T_{22}
44C20x20=44C21x21{ }^{44} C_{20} x^{20}={ }^{44} C_{21} x^{21}
x=44C2044C21\Rightarrow x=\frac{{ }^{44} C_{20}}{{ }^{44} C_{21}}
x=44!20!×24!44!21!×23!=21!×23!20!×24!\Rightarrow x=\frac{\frac{44 !}{20 ! \times 24 !}}{\frac{44 !}{21 ! \times 23 !}}=\frac{21 ! \times 23 !}{20 ! \times 24 !}
=2124=78=\frac{21}{24}=\frac{7}{8}