Question

Two successive terms in the expansion of $(1+x{)}^{24}$ whose coefficients are in the ratio 4 : 1 are

• A. ${18}^{th},{19}^{th}$
• B. ${19}^{th},{20}^{th}$
• C. ${20}^{th},{21}^{st}$
• D. ${21}^{st},{22}^{nd}$

Answer 1

Answer: (C)

${20}^{th},{21}^{st}$

Using Binomial theorem,

$(1+x{)}^{24}={\sum }_{n=0}^{24}{}^{24}{C}_n{x}^n$

$T_{x+1}{=}^{24}{C}_x{x}^n$

Let $T_x,T_{x+1}$ be the two successive terms of $(1+x{)}^{24}$ whose coefficients are in the ratio $4:1$

$\Rightarrow \frac{{}^{24}{C}_{n-1}}{{}^{24}{C}_n}=\frac{4}{1}$

${\Rightarrow }^{24}{C}_{n-1}=4^{24}{C}_n$

$\Rightarrow \frac{\left(24\right)!}{(n-1)!(24-n+1)!}=4.\frac{\left(24\right)!}{n!(24-n)!}$

$\Rightarrow [using{}^n{e}_n=\frac{n!}{n!(n-r)!}]$

$\Rightarrow \frac{1}{(r-1)!(24-n+1)(24-n)!}=\frac{4}{r(r-1)!(24-r)!}$

$\Rightarrow \frac{1}{25-r}=\frac{4}{r}$

$\Rightarrow r=100-4r$

$\Rightarrow r+4r=100$

$\Rightarrow 5r=100$

$\Rightarrow r=20$.

$\therefore$ the two successive terms are ${20}^{th},{21}^{st}$