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.

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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, $T_{n} = (- 1)^{n} [n^{3} + (n - 1)^{3}] = (- 1)^{n} n^{3} - (- 1)^{n - 1} (n - 1)^{3} = V_{n} - V_{n - 1}$
$\Rightarrow \sum_{n = 1}^{102} T_{n} = \sum_{n = 1}^{102} (V_{n} - V_{n - 1}) = V_{102} - V_{0} = (102)^{3}$
$[\frac{\sum_{n = 1}^{102} T_{n}}{520200}] = 2$