Question

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

Answer 1

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

$$\begin{gathered} T_n={}^{44}C_{n-1}{\left(x\right)}^{44-\left(n-1\right)}{\left(1\right)}^{n-1} \\ \Rightarrow T_{21}={}^{44}C_{20}{\left(x\right)}^{44-\left(21-1\right)}{\left(1\right)}^{21-1}={}^{44}C_{20}{\left(x\right)}^{24} \end{gathered}$$

And $\Rightarrow T_{22}={}^{44}C_{21}{\left(x\right)}^{44-\left(22-1\right)}{\left(1\right)}^{22-1}={}^{44}C_{21}{\left(x\right)}^{23}$

Now,

$$\begin{gathered} T_{21}=T_{22} \\ \Rightarrow {}^{44}C_{20}{\left(x\right)}^{24}={}^{44}C_{21}{\left(x\right)}^{23} \\ \Rightarrow x=\frac{{}^{44}C_{21}}{{}^{44}C_{20}}=\frac{\left(44!\right)\left(21!\right)\left(23!\right)}{\left(44!\right)\left(20!\right)\left(24!\right)}=\frac{7}{8} \end{gathered}$$