Let An,(n∈N)be a matrix of order (2n−1)×(2n−1),such that aij=0∀i≠jand aij=n2+i+1−2n,∀i=jwhere aij denotes the element of ith row and jth column of An. Let Tn=-1n·sum of all the elements of An Find the value of∑n=1102 Tn520200, where [.] represents the greatest integer function.

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Let $A_{n}, (n \in N)$ be a matrix of order $(2 n - 1) \times (2 n - 1)$ ,such that $a_{ij} = 0 \forall i \neq j$ and $a_{ij} = n^{2} + i + 1 - 2 n, \forall i = j$ where $a_{ij}$ denotes the element of i^th row and j^th column of $A_{n}$ .
Let $T_{n} = {(- 1)}^{n} \cdot (s u m o f a l l t h e e l e m e n t s o f A_{n})$
Find the value of $[\frac{\sum_{n = 1}^{102} T_{n}}{520200}]$ , where [.] represents the greatest integer function.

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answer is 2.

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Detailed Solution

$a_{ij} = 0 \forall i \neq j$ and $a_{ij} = (n - 1)^{2} + i \forall i = j$
Sum of all the element of $A_{n} = \sum_{i = 1}^{2 n - 1} [(n - 1)^{2} + i]$
$= (2 n - 1) (n - 1)^{2} + (2 n - 1) n = 2 n^{3} - 3 n^{2} + 3 n - 1 = n^{3} + (n - 1)^{3}$
So, $T_{n} = (- 1)^{n} [n^{3} + (n - 1)^{3}] = (- 1)^{n} n^{3} - (- 1)^{n - 1} (n - 1)^{3} = V_{n} - V_{n - 1}$
$\Rightarrow \sum_{n = 1}^{102} T_{n} = \sum_{n = 1}^{102} (V_{n} - V_{n - 1}) = V_{102} - V_{0} = (102)^{3}$
$[\frac{\sum_{n = 1}^{102} T_{n}}{520200}] = 2$