Question

Let $t_n$ denote the $n^{th}$ term in a binomial expansion. If $ \frac{t_{6}}{t_{5}}$ in the expansion of $(a+ b)^{n+4}$ and $ \frac{t_{5}}{t_{4}}$ in the expansion of $(a + b)^n$ are equal, then $n$ is

• A. 9
• B. 11
• C. 13
• D. 15

Answer 1

The Correct Option is D

$t_{6}$ and $t_{5}$ in the expansion of $(a+b)^{n+4}$ is
$t_{5}=t_{4+1}={ }^{n+4} C_{4} a^{n+4-4} b^{4}={ }^{n+4} C_{4} a^{n} \cdot b^{4}$
and $t_{6}=t_{5+1}={ }^{n+4} C_{5} a^{n+4-5} b^{5}={ }^{n+4} C_{5} a^{n-1} b^{5}$
$\therefore\, \frac{t_{6}}{t_{5}}=\frac{{n+4} C_{5} \cdot a^{n-1} \cdot b^{5}}{{ }^{n+4} C_{4} \cdot a^{n} \cdot b^{4}}$
$=\frac{{ }^{n+4} C_{5}}{{ }^{n+4} C_{4}}\left(\frac{b}{a}\right)=\frac{n}{5}\left(\frac{b}{a}\right)\,\,\,\,\,\,\,\dots(i)$
Now, $t_{5}$ and $t_{4}$ in the expansion of $(a+b)^{n}$ is
$ t_{5}=t_{4+1}={ }^{n} C_{4} \cdot a^{n-4} \cdot b^{4} $
and $t_{4}=t_{3+1}={ }^{n} C_{3} \cdot \hat{a}^{n-3} \cdot b^{3} $
$\therefore \,\frac{t_{5}}{t_{4}}=\frac{{ }^{n} C_{4} \cdot a^{n-4} \cdot b^{4}}{{ }^{n} C_{3} \cdot a^{n-3} \cdot b^{3}}=\frac{n-3}{4} \cdot\left(\frac{b}{a}\right)\,\,\,\,\,\,\,\,\dots(ii)$
On equating Eqs. (i) and (ii), we get
$\frac{n}{5}\left(\frac{b}{a}\right)=\frac{n-3}{4}\left(\frac{b}{a}\right)$
$\Rightarrow \, 4 n=5 n-15$
$\Rightarrow\, n=15$