Question

The set of natural number N is partitioned into arrays of rows and columns in the form of matrices as $m_1=\left(1\right),m_2=\left(\begin{array}{cc}2& 3\\ 4& 5\end{array}\right),m_3=\left(\begin{array}{ccc}6& 7& 8\\ 9& 10& 11\\ 12& 13& 14\end{array}\right),.....,m_n=\left(\begin{array}{ccc}.& .& .\\ .& .& .\\ .& .& .\end{array}\right)$ and so on then

• A. the first term in $m_{10}$ matrix is $286$
• B. the first term in $m_{10}$ matrix is $386$
• C. sum of the elements of the diagonal in $m_{10}$ is $3355$
• D. sum of the elements of the diagonal in $m_{10}$ is $4455$

Answer 1

Answer: (A), (C)

The correct options are
A sum of the elements of the diagonal in $m_{10}$ is $3355$
C the first term in $m_{10}$ matrix is $286$

Sum of number of elements of $m_{n-1}$ matrices will give you the last element of $m_{n-1}$ matrix
So, last element of $m_{n-1}$ matrix is = $1^2+2^2+3^2+...+{(n-1)}^2=\frac{n(n-1)(2n-1)}{6}$
So, first element of $m_{10}$ matrix = $\frac{10\times 9\times 19}{6}+1=286$
Common difference in the diagonal elements of $m_n$ matrix is n + 1, so in $m_{10}$, common different = 11
Sum of diagonal elements = $\frac{10}{2}[2\times 286+(10-1)\times 11]=5(572+99)=5\times 671=3355$