Question

Let $A_n(n\in N)$ be a square matrix of order $(2n-1)\times (2n-1)$ such that $a_{ij}=0\forall i\ne j$ and $a_{ij}=n^2+i+1-2n\forall i=j$ where $a_{ij}$ denotes the element of $i^{th}$ row and $j^{th}$ column of $A_n.$ Let $T_n=(-1{)}^n\times (\text{sum of all the elements of}{A}_n).$ Let $S=\sum _{n=1}^{102}{T}_n$ , then the value of $S^{\frac{1}{3}}$,

Answer 1

$a_{ij}=0,\forall i\ne j$
$a_{ij}=(n-1{)}^2+i,\forall i=j$

Sum of all the elements of $A_n$
$=\sum _{i=1}^{2n-1}[(n-1{)}^2+i]$
$=(2n-1)(n-1{)}^2+(2n-1)n$
$=2n^3-3n^2+3n-1$
$=n^3+(n-1{)}^3$

$T_n=(-1{)}^n[n^3+(n-1{)}^3]$
$=(-1{)}^n{n}^3-(-1{)}^{n-1}(n-1{)}^3$
$=V_n-V_{n-1}$
$\sum _{n=1}^{102}{T}_n=\sum _{n=1}^{102}(V_n-V_{n-1})=V_{102}-V_0=(102{)}^3$