Question

The set of natural numbers is divided into arrays of rows and columns in the form of matrices as
$A_1=\left(1\right),A_2=\left(\begin{array}{cc}2& 3\\ 4& 5\end{array}\right),A_3=\left(\begin{array}{ccc}6& 7& 8\\ 9& 10& 11\\ 12& 13& 14\end{array}\right)$... so on.
Find the value of $T_r\left(A_{10}\right)$.
[Note: $T_r\left(A\right)$ denotes sum of diagonal elements of $A$.]

• A. $3355$
• B. $3434$
• C. $5533$
• D. None of these

Answer 1

The correct option is A $3355$

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

$1^1+2^2+3^2+\cdots +(n-1{)}^2+1=\sum _{i=1}^{n-1}{i}^2+1=\frac{1}{6}n(n-1)(2n-1)+1=k$ (say)
Now the diagonal elements of $A_n$ will be $k,k+(n+1),k+2(n+1)\cdots k+(n+1)r,\cdots k+(n+1)(n-1)$
Hence,

$T_r\left(A_n\right)=k+k+(n+1)+k+2(n+1)+\cdots +k+(n+1)r+\cdots +k+(n+1)(n-1)$
$=nk+(n+1)\sum _{r=1}^{n-1}r=nk+\frac{(n-1)n(n+1)}{2}$

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

$=\frac{n}{6}(2n^3+n+3)$
Substitute, $n=10$

$Tr\left(A_{10}\right)=3355$

Hence, option A.