# if the 21st and 22nd terms in the expansion of (1+x)^44 are equal then find the value of x.

The nth term of the binomial expansion ${\left(1+x\right)}^{44}$ is given as:

${T}_{n}={}^{44}C_{n-1}{\left(x\right)}^{44-\left(n-1\right)}{\left(1\right)}^{n-1}\phantom{\rule{0ex}{0ex}}⇒{T}_{21}={}^{44}C_{20}{\left(x\right)}^{44-\left(21-1\right)}{\left(1\right)}^{21-1}={}^{44}C_{20}{\left(x\right)}^{24}$

And $⇒{T}_{22}={}^{44}C_{21}{\left(x\right)}^{44-\left(22-1\right)}{\left(1\right)}^{22-1}={}^{44}C_{21}{\left(x\right)}^{23}$

Now,

${T}_{21}={T}_{22}\phantom{\rule{0ex}{0ex}}⇒{}^{44}C_{20}{\left(x\right)}^{24}={}^{44}C_{21}{\left(x\right)}^{23}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{}^{44}C_{21}}{{}^{44}C_{20}}=\frac{\left(44!\right)\left(21!\right)\left(23!\right)}{\left(44!\right)\left(20!\right)\left(24!\right)}=\frac{7}{8}$

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