Q. If $D_{k} = ∣ ∣ 1 2 k 2 k - 1 n n^{2} + n + 1 n^{2} n n^{2} + n n^{2} + n + 1 ∣ ∣$ and $k = 1 \sum n D_{k} = 56$ , then $n$ equals to

3811 150 NTA AbhyasNTA Abhyas 2022 Report Error

A

$4$

B

$8$

C

$6$

D

None of these

Solution:

It is given that $Σ_{k = 1}^{n} D_{k} = 56$
where, $D_{k} = ∣ ∣ 1 2 k 2 k - 1 n n^{2} + n + 1 n^{2} n n^{2} + n n^{2} + n + 1 ∣ ∣$
$\Rightarrow ∣ ∣ Σ_{k = 1}^{n} 1 Σ_{k = 1}^{n} 2 k Σ_{k = 1}^{n} (2 k - 1) n n^{2} + n + 1 n^{2} n n^{2} + n n^{2} + n + 1 ∣ ∣ = 56$
$\Rightarrow ∣ ∣ n n (n + 1) n^{2} n n^{2} + n + 1 n^{2} n n^{2} + n n^{2} + n + 1 ∣ ∣ = 56$
$C_{2} \to C_{2} - C_{1}$
$C_{3} \to C_{3} - C_{1}$
$∣ ∣ n n^{2} + n n^{2} 010 00 n + 1 ∣ ∣ = 56$
$\Rightarrow n (n + 1) = 56$
$\Rightarrow n = 7$