Question

If $\frac{t_2}{t_3}$ in the expansion of $(a+b{)}^n$ and $\frac{t_3}{t_4}$ in the expansion of $(a+b{)}^{n+3}$ are equal, then find the value of $n$

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is A $3$

using binomial theorem,

$(a+b{)}^n=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)a^{n-k}{b}^k$, where $T_{n+1}=\left(\begin{array}{c}n\\ k\end{array}\right)a^{n-r}{b}^r$

$(a+b{)}^{n+b}=\sum _{k=0}^{n+3}\left(\begin{array}{c}n+3\\ k\end{array}\right)a^{n+3-k}{b}^k$

In $(a+b{)}^n$,

The second term of $(a+b{)}^n,t_2=\left(\begin{array}{c}n\\ 1\end{array}\right)a^{n-1}{b}^1$

$=n{a}^{n-1}b$.

The third term of $(a+b{)}^n,t_3=\left(\begin{array}{c}n\\ 2\end{array}\right)a^{n-2}{b}^2=\frac{n!}{2!(n-2)!}{a}^{n-2}{b}^2$

$=\frac{n(n-1)(n-2)!}{2(n-2)!}{a}^{n-2}{b}^2$

$=\frac{n(n-1)}{2}{a}^{n-2}{b}^2$

$\therefore \frac{t_2}{t_3}=\frac{n{a}^{n-1}b}{\frac{n(n-1)}{l}{a}^{n-2}{b}^2}=\frac{2a}{(n-1)b}$.......(1)

In $(a+b{)}^{n+3}$,

The third term, $t_3=\left(\begin{array}{c}n+3\\ 2\end{array}\right)a^{n+3-2}{b}^2=\frac{(n+3)!}{2!(n+3-2)!}{a}^{n+1}{b}^2$

$=\frac{(n+3)(n+2)(n+1)!}{2(n+1)!}{a}^{n+1}{b}^2$

$=\frac{(n+3)(n+2)}{2}{a}^{n+1}{b}^2$

The forth term, $t_4=\left(\begin{array}{c}n+3\\ 3\end{array}\right)a^{n+3-3}{b}^3=\frac{(n+3)!}{3!(n+3-3)!}{a}^n{b}^3$

$=\frac{(n+3)(n+2)(n+1)n!}{6n!}{a}^n{b}^3$

$=\frac{(n+3)(n+2)(n+1)}{6}{a}^n{b}^3$

$\therefore \frac{t_3}{t_4}=\frac{\frac{(n+3)(n+2)a^{n+1}{b}^2}{2}}{\frac{(n+3)(n+2)(n+1)}{6}{a}^n{b}^3}=\frac{3a}{(n+1)b}$.......(2)

As per question, (1) and (2) are equal

ie $\frac{2a}{(n-1)b}=\frac{3a}{(n+1)b}$

$\Rightarrow 2(n+1)=3(n-1)$

$\Rightarrow 2n+2=3n-1$

$\Rightarrow 2+1=3n-2n$

$\Rightarrow n=3$.