Question

Let $A=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)$. Then for positive integer $n,A^n$ is

• A. $\left(\begin{array}{ccc}1& n& n^2\\ 0& n^2& n\\ 0& 0& n\end{array}\right)$
• B. $\left(\begin{array}{ccc}1& n& n\left(\frac{n+1}{2}\right)\\ 0& 1& n\\ 0& 0& 1\end{array}\right)$
• C. $\left(\begin{array}{ccc}1& n^2& n\\ 0& n& n^2\\ 0& 0& n^2\end{array}\right)$
• D. $\left(\begin{array}{ccc}1& n& 2n-1\\ 0& \frac{n+1}{2}& n^2\\ 0& 0& \frac{n+1}{2}\end{array}\right)$

Answer 1

Answer: (B)

$\left(\begin{array}{ccc}1& n& n\left(\frac{n+1}{2}\right)\\ 0& 1& n\\ 0& 0& 1\end{array}\right)$

$A=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)$
$A^2=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right)$
$A^3=\left(\begin{array}{ccc}1& 3& 6\\ 0& 1& 3\\ 0& 0& 1\end{array}\right)$
There using mathematical induction we get that
$A^n=\left(\begin{array}{ccc}1& n& \frac{n(n+1)}{2}\\ 0& 1& n\\ 0& 0& 1\end{array}\right)$