Here we see that An has n2 terms. Hence total number of terms upto A9 is 1+22+32+…+92 =9(9+1)(2×9+1)6=285 using formula n(n+1)(2n+1)6 Hence A10 will start with a286.
A10=⎡⎢
⎢
⎢
⎢
⎢
⎢⎣a286a287⋯a295a296a297⋯a305a306a307⋯a315⋅⋅⋅⋅⋅⋅⋅⋅⎤⎥
⎥
⎥
⎥
⎥
⎥⎦10×10
This A10 is a 10×10 matrix.
Trace of A10 is sum of diagonal elements of A10
Following the difference of 11 in term number we can extrapolate all the 10 diagonal terrms.
Tr(A)=a286+a297+a308+a319+a330+a341+a352+a362+a363+a374+a385 =[log2286]+[log2297]+[log2308]+[log2319]+[log2330]+[log2341]+[log2352]+ [log2363]+[log2374]+[log2385] Now, 28=256 and 29=512
⇒log2x∈(8,9)∀x∈(256,512)
Since, ar=[log2r] and here 256<r<512⇒log2r∈(8,9)⇒[log2r]=8
Hence tr(A)=8+8+8+⋯upto 10 times =80