Question

The set of natural numbers $N$ is partitioned into arrays of rows and columns in the form of matrices as $M_1=\left(1\right),M_2=\left(\begin{array}{cc}2& 3\\ 4& 5\end{array}\right),M_3=\left(\begin{array}{ccc}6& 7& 8\\ 9& 10& 11\\ 12& 13& 14\end{array}\right),...,M_n=\left(\begin{array}{}\end{array}\right)$ and so on.

Find the sum of the elements of the diagonal in $M_n$ for $n=6$

Answer 1

Here clearly it can be observed that $A_1$ contain $1^2=1$ elements.
$A_2$ contains $2^2=4$ elements. $A_3$ contains $3^2=9$
elements and so on. Therefore $A_n$ will contain $n^2$ elements.
Thus
the first element of $A_n$ will be $1^1+2^2+3^2+...........(n-1{)}^2+1=\sum _{i=1}^{n-1}{i}^2+1=\frac{1}{6}n(n-1)(2n-1)+1=k$ (say)

Now the diagonal elements of $A_n$ will be

$k,k+(n+1),k+2(n+1)..............k+(n+1)r,.......k+(n+1)(n-1)$

Hence $T_r\left(A_n\right)=k+k+(n+1)+k+2(n+1)+..............+k+(n+1)r+.......+k+(n+1)(n-1)$

$=nk+(n+1)\sum _{r=1}^{n-1}=nk+\frac{(n-1)n(n+1)}{2}$

$=\frac{1}{6}{n}^2(n-1)(2n-1)+n+\frac{(n-1)n(n+1)}{2}=\frac{n}{6}(2n^3+n+3)$

$\therefore Tr\left(A_6\right)=441$