Question

The ${21}^{st}$ and ${22}^{nd}$ terms in the expansion of $(1+x{)}^{44}$ are equal. Then $x=$

• A. 8/7
• B. 7/8
• C. 7
• D. 8

Answer 1

Answer: (B)

7/8

It is given that $T_{21}=T_{22}$
$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question, we get

$\Rightarrow {}^{44}{C}_{20}{x}^{20}={}^{44}{C}_{21}{x}^{21}$

$\Rightarrow \frac{44!}{24!20!}=\frac{44!}{23!21!}x$

$\Rightarrow x=\frac{21}{24}$

$\Rightarrow x=\frac{7}{8}$