Question

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

• A. $\frac {7} {8}$
• B. $\frac {8} {7}$
• C. $\frac {21} {22}$
• D. $\frac {23} {24}$

Answer 1

Answer: (A)

$T_{21} = T_{20+1} = \,{}^{44}c_{20}x^{20}$
$T_{22} = T_{21+1} = \,{}^{44}c_{21}\left(-x\right)^{21} = - \,{}^{44}c_{21}x^{21}$
$\therefore \,{}^{44}c_{20}x^{20} = - \,{}^{44}c_{21}x^{21}$
$\therefore x = \frac{^{44}c_{20}}{^{44}c_{21}} = \frac{44\,!}{24\,!\,20\,!} \times \frac{21\,!\,3!}{44\,!}$
$= \frac{21\cdot20\,!\,23\,!}{20\,!\,24\cdot23\,!} = -\frac{21}{24} = -\frac{7}{8}$