If 21^{text{st}} and 22^{text{nd}} terms in t | vedclass

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(D) The general term in the expansion of $(1+x)^n$ is given by $T_{r+1} = {}^nC_r x^r$.For the expansion of $(1+x)^{44}$,the $21^{	ext{st}}$ term is

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$ \frac{7}{8} $

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(D) The general term in the expansion of $(1+x)^n$ is given by $T_{r+1} = {}^nC_r x^r$.For the expansion of $(1+x)^{44}$,the $21^{	ext{st}}$ term is $T_{21} = T_{20+1} = {}^{44}C_{20} x^{20}$.The $22^{	ext{nd}}$ term is $T_{22} = T_{21+1} = {}^{44}C_{21} x^{21}$.Given that $T_{21} = T_{22}$,we have ${}^{44}C_{20} x^{20} = {}^{44}C_{21} x^{21}$.Dividing both sides by $x^{20}$ (assuming $x 
eq 0$),we get $x = \frac{{}^{44}C_{20}}{{}^{44}C_{21}}$.Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:$x = \frac{44!}{20!24!} 	imes \frac{21!23!}{44!} = \frac{21!}{20!} 	imes \frac{23!}{24!} = \frac{21}{1} 	imes \frac{1}{24} = \frac{21}{24} = \frac{7}{8}$.

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

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