Question

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

Answer 1

Here we see that $A_n$ has $n^2$ terms.

Hence total number of terms upto $A_9$ is $1+2^2+3^2+...+9^2$

$=\frac{9(9+1)(2\times 9+1)}{6}=285$ using formula $\frac{n(n+1)(2n+1)}{6}$

Hence $A_{10}$ will start with $a_{286}$

$A_{10}=\left[\begin{array}{ccccccccc}{a}_{286}& a_{287}& .& .& .& .& .& .& a_{295}\\ a_{296}& a_{297}& a_{298}& .& .& .& .& .& a_{305}\\ a_{306}& a_{307}& a_{308}& .& .& .& .& .& a_{315}\\ .& .& .& .& .& & & & \\ .& .& .& .& .& & & & \\ .& .& .& .& .& & & & \end{array}\right]$

This $A_{1}0$ 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 no. we can extrapolate all the 10 diagonal terrms.

$Tr\left(A\right)=a_{286}+a_{297}+a_{308}+a_{319}+a_{330}+a_{341}+a_{352}+a_{363}+a_{374}+a_{385}$

$=\left[{\log }_{2}286\right]+\left[{\log }_{2}297\right]+\left[{\log }_{2}308\right]+\left[{\log }_{2}319\right]+\left[{\log }_{2}330\right]+\left[{\log }_{2}341\right]+\left[{\log }_{3}52\right]+\left[{\log }_{2}363\right]+\left[{\log }_{2}374\right]+\left[{\log }_{2}385\right]$

Now, $2^8=256$ and $2^9=512$

$⟹{\log }_{2}x\in (8,9)\forall x\in (256,512)$

Since $a_r=\left[{\log }_{2}r\right]$ and here $256<r<512$

$⟹{\log }_{2}r\in (8,9)$

$⟹\left[{\log }_{2}r\right]=8$

Hence $Tr\left(A\right)=8+8+\mathrm{8...}$upto 10 times $=8\times 10$ $

$Tr\left(A\right)=80$