If A=aijn×n such that aij=(i+j)2 then trace of A is

aij=(i+j)2 A=a11+a22+a33+……+ann =22+42+62+……+(2n)2 =4(1+4+9+……+n2) =4n(n+1)(2n+1)6 =23n(n+1)(2n+1)

If A=aijn×n such that aij=(i+j)2 then trace of A is

aij=(i+j)2 A=a11+a22+a33+……+ann =22+42+62+……+(2n)2 =4(1+4+9+……+n2) =4n(n+1)(2n+1)6 =23n(n+1)(2n+1)

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If $A = {[a_{ij}]}_{n \times n}$ such that $a_{ij} = {(i + j)}^{2}$ then trace of A is

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$\frac{1}{3} n (n + 1) (2 n + 1)$

$\frac{2}{3} n (n - 1) (2 n - 1)$

$\frac{2}{3} n (n + 1) (2 n + 1)$

$\frac{1}{3} n (n - 1) (2 n - 1)$

answer is C.

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Detailed Solution

$\begin{array}{rcl} a_{ij} & = & {(i + j)}^{2} \\ A & = & a_{11} + a_{22} + a_{33} + \dots \dots + a_{nn} \\ = & 2^{2} + 4^{2} + 6^{2} + \dots \dots + {(2 n)}^{2} \\ = & 4 (1 + 4 + 9 + \dots \dots + n^{2}) \\ = & 4 \frac{n (n + 1) (2 n + 1)}{6} \\ = & \frac{2}{3} n (n + 1) (2 n + 1) \end{array}$