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Question

Let A1=[a1]A2=[a2a3a4a5]A3=a6a7a8a9a10a11a12a13a14An=[] where ar=[log2r]
([] denotes the greatest integer). Then trace of A10 is

A
80.00
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B
80
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C
80.0
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Solution

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=⎢ ⎢ ⎢ ⎢ ⎢ ⎢a286a287a295a296a297a305a306a307a315⎥ ⎥ ⎥ ⎥ ⎥ ⎥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<512log2r(8,9)[log2r]=8
Hence tr(A)=8+8+8+upto 10 times =80

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