Question

If the $20th$ and $21s$t terms in the expansion of $(1+x{)}^{40}$ are equal, then the value of x is

• A. $\frac{20}{21}$
• B. $\frac{21}{20}$
• C. $25$
• D. $\frac{1}{25}$

Answer 1

The correct option is A $\frac{20}{21}$

Given,

$(1+x{)}^{40}$

General term

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$T_{r+1}={}^{40}{C}_r\left(1{)}^{40-r}\right(x{)}^r$

Given $20$th and $21$st terms in the expansion are equal.

$T_{20}=T_{19+1}={}^{40}{C}_{19}\left(1{)}^{40-19}\right(x{)}^{19}$

$T_{20}={}^{40}{C}_{19}{x}^{19}$

$T_{21}=T_{20+1}={}^{40}{C}_{20}\left(1{)}^{40-20}\right(x{)}^{20}$

${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$T_{21}={}^{40}{C}_{20}{x}^{20}$

$T_{20}=T_{21}$

${}^{40}{C}_{19}{x}^{19}={}^{40}{C}_{20}{x}^{20}$

$\frac{40!}{19!\times 21!}{x}^{19}=\frac{40!}{20!\times 20!}{x}^{20}$

$\frac{x^{19}}{21}=\frac{x^{20}}{20}$

$\frac{20}{21}=x$

Hence correct option is A.