Question

In the expansion of $(a+b{)}^n$ if two consective terms are equal, then $\frac{(n+1)b}{a+b}$ and $\frac{(n+1)a}{a+b}$ are:

• A. Integers
• B. Rational Numbers but not integers
• C. Irrational numbers
• D. Cannot be determined.

Answer 1

The correct option is A Integers

It is given that $T_r=T_{r+1}$
Therefore ${}^n{C}_{r-1}{a}^{n-(r-1)}{b}^{r-1}={}^n{C}_r{a}^{n-r}{b}^r$
${}^n{C}_{r-1}a={}^n{C}_{r}b$
$\frac{n!a}{(n-(r-1\left)\right)!(r-1)!}=\frac{n!b}{(n-r)!r!}$
$\frac{a}{n+1-r}=\frac{b}{r}$
$ar=bn+b-br$
$ar+br=b(n+1)$
$r=\frac{b(n+1)}{a+b}$ ...(i)
Now $(a+b{)}^n$ can also be written as $(b+a{)}^n$
It is given that $T_r=T_{r+1}$
Therefore ${}^n{C}_{r-1}{b}^{n-(r-1)}{a}^{r-1}={}^n{C}_r{b}^{n-r}{a}^r$
${}^n{C}_{r-1}b={}^n{C}_{r}a$
$\frac{n!b}{(n-(r-1\left)\right)!(r-1)!}=\frac{n!a}{(n-r)!r!}$
$\frac{b}{n+1-r}=\frac{a}{r}$
$br=an+a-ar$
$ar+br=a(n+1)$
$r=\frac{a(n+1)}{a+b}$ ...(ii)
Since '$r$' is the term number, hence it can only be a positive integer. Hence $\frac{a(n+1)}{a+b}$ and $\frac{b(n+1)}{a+b}$ are positive integers only.