Let $A=\left(a_{ij}\right)$ be an $n\times n$ matrix defined by $a_{ij}=\left\{\begin{array}{l}{k}^i, \forall i=j\\ 0, \text{otherwise}\end{array}\right.$. If $m=$ trace of $A$ and $\underset{k\to 1}{\lim }\frac{n-m}{1-k}=171$ then the value of $n$ is

Given $a_{ij}=\left\{\begin{array}{l}{k}^i, \forall i=j\\ 0, \mathrm{otherwise}\end{array}\right.$

i.e. $m=k^1+k^2+k^3+....k^n$

$=\frac{k\left(1-k^n\right)}{1-k}$

Now, $\underset{k\to 1}{\lim }\frac{n-m}{1-k}=171$

i.e. $\underset{k\to 1}{\lim }\frac{{\displaystyle n\left(1-k\right)-k\left(1-k^n\right)}}{{\left(1-k\right)}^2}=171$

$\Rightarrow \underset{k\to 1}{\lim }\frac{{\displaystyle n-k\left(1+k+...k^{n-1}\right)}}{\left(1-k\right)}=171$

Applying L'Hospital rule, we get

$\underset{k\to 1}{\lim }\frac{{\displaystyle -1-2k-...-n{k}^{n-1}}}{-1}=171$

$\Rightarrow 1+2+..n=171\Rightarrow \frac{n\left(n+1\right)}{2}=171$

Hence, $n=18$