Question

In the expansion of ${(a\sqrt{a}+\frac{1}{a^4})}^n$, the coefficient in the second term exceeds by $44$ the coefficient in the first term. Find $n$.

• A. $20$
• B. $25$
• C. $35$
• D. $45$

Answer 1

The correct option is D $45$

Given, ${(a\sqrt{a}+\frac{1}{a^4})}^n$

$\Rightarrow coeff\left(T_2\right)=44+coeff\left(T_1\right)$

To find: $n$

Now, $T_2=T_{1+1}={}^n{C}_1{\left(a\sqrt{a}\right)}^{n-1}\left(\frac{1}{a^4}\right)={}^n{C}_1{a}^{3/2(n-1)-4}$

$coeff\left(T_2\right)={}^n{C}_1$

$T_1=T_{0+1}={}^n{C}_0{\left(a\sqrt{a}\right)}^{n-0}{\left(\frac{1}{a^4}\right)}^0$

$coeff\left(T_1\right)={}^n{C}_0$

$\therefore {}^n{C}_1=44+{}^n{C}_0$

$\Rightarrow \frac{n!}{(n-1)!}=44+1$

$\Rightarrow n=45$