Question

Let $A_1=\left[a_1\right]A_2=\left[\begin{array}{cc}{a}_2& a_3\\ a_4& a_5\end{array}\right]A_3=\left[\begin{array}{ccc}{a}_6& a_7& a_8\\ a_9& a_{10}& a_{11}\\ a_{12}& a_{13}& a_{14}\end{array}\right]\cdots \cdots A_n=[\cdots ]\text{where}{a}_r=\left[{\log }_{2}r\right]$
$\left(\right[\cdot ]$ denotes the greatest integer$)$. Then trace of $A_{10}$ is

• A. 80.00
• B. 80
• C. 80.0

Answer 1

Answer: (A), (B), (C)

Here we see that $A_n$ has $n^2$ terms. Hence total number of terms upto $A_9$ is $1+2^2+3^2+\dots +9^2$ $=\frac{9(9+1)(2\times 9+1)}{6}=285$ using formula $\frac{n(n+1)(2n+1)}{6}$ Hence $A_{10}$ will start with $a_{286}$.
$A_{10}={\left[\begin{array}{ccccc}{a}_{286}& a_{287}& \cdots & a_{295}& \\ a_{296}& a_{297}& \cdots & a_{305}& \\ a_{306}& a_{307}& \cdots & a_{315}& \\ \cdot & \cdot & \cdot & \cdot & \\ \cdot & \cdot & \cdot & \cdot & \end{array}\right]}_{10\times 10}$
This $A_{10}$ is a $10\times 10$ matrix.
Trace of $A_{10}$ is sum of diagonal elements of $A_{10}$
Following the difference of $11$ in term number we can extrapolate all the $10$ diagonal terrms.
$Tr\left(A\right)=a_{286}+a_{297}+a_{308}+a_{319}+a_{330}+a_{341}+a_{352}+a_{362}+a_{363}+a_{374}+a_{385}$ $=\left[{\log }_{2}286\right]+\left[{\log }_{2}297\right]+\left[{\log }_{2}308\right]+\left[{\log }_{2}319\right]+\left[{\log }_{2}330\right]+\left[{\log }_{2}341\right]+\left[{\log }_{2}352\right]+$ $\left[{\log }_{2}363\right]+\left[{\log }_{2}374\right]+\left[{\log }_{2}385\right]$ Now, $2^8=256$ and $2^9=512$
$\Rightarrow {\log }_{2}x\in (8,9)\forall x\in (256,512)$
Since, $a_r=\left[{\log }_{2}r\right]$ and here $256<r<512\Rightarrow {\log }_{2}r\in (8,9)\Rightarrow \left[{\log }_{2}r\right]=8$
Hence $tr\left(A\right)=8+8+8+\cdots$upto $10$ times $=80$