Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 60 math

$A^{n}=\left[\begin{array}{ccc}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right]$ for every integer n.

Hint: We use the principle of mathematical induction.

Given:

$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$

Prove:

$A^{n}=\left[\begin{array}{ccc}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right]$ for every positive integer n.                 …(i)

Solution:

step 1: put n=1 in eqn(i)

$A^{\mathbf{1}}=\left[\begin{array}{ccc}1 & 1 & \frac{1(1+1)}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\\\\ A^{\mathbf{1}}=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$

So, $A^n$ is true for n=1

Step 2 : let  $A^n$ be true for n=k, so

$A^{k}=\left[\begin{array}{ccc}1 & k & \frac{k(k+1)}{2} \\ 0 & 1 & k \\ 0 & 0 & 1\end{array}\right]$

Step 3:  we will prove  $A^n$ that will be true for n=k+1

Now,$\\A^{k+\mathbf{1}}=A^{k} A\) \(=\left[\begin{array}{ccc}1 & k & \frac{k(k+1)}{2} \\ 0 & 1 & k \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1+0+0 & 1+k+0 & 1+k+\frac{k(k+1)}{2} \\ 0+0+0 & 0+1+0 & 0+1+k \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1 & k+1 & \frac{(k+1)(k+2)}{2} \\ 0 & 1 & (k+1) \\ 0 & 0 & 1\end{array}\right]$

Hence, $A^n$ is true for n=k+1 wherever it is true for n=k

So, by principle of mathematical induction $A^n$  is true for all positive integers n.