If ∆r = r2r-13r-2n2n-1a12n(n-1)(n-1)212(n-1)(3n+4) then the value of ∑r=1n-1 ∆r

∆r = r2r-13r-2n2n-1a12n(n-1)(n-1)212(n-1)(3n+4)∑r=1n-1 ∆r=∑r=1n-1 r∑r=1n-1 (2r-1)∑r=1n-1 (3r-2)n2n-1a12n(n-1)(n-1)212(n-1) (3n+4)∑r=1n-1∆r = 12n(n-1)(n-1)212(n-1)(3n+4)n2n-1a12n(n-1)(n-1)212(n-1)(3n+4)=0it is clearly independent of both 'a' and 'n'

If ∆r = r2r-13r-2n2n-1a12n(n-1)(n-1)212(n-1)(3n+4) then the value of ∑r=1n-1 ∆r

∆r = r2r-13r-2n2n-1a12n(n-1)(n-1)212(n-1)(3n+4)∑r=1n-1 ∆r=∑r=1n-1 r∑r=1n-1 (2r-1)∑r=1n-1 (3r-2)n2n-1a12n(n-1)(n-1)212(n-1) (3n+4)∑r=1n-1∆r = 12n(n-1)(n-1)212(n-1)(3n+4)n2n-1a12n(n-1)(n-1)212(n-1)(3n+4)=0it is clearly independent of both 'a' and 'n'

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If $∆_{r} = |\begin{matrix} r & 2 r - 1 & 3 r - 2 \\ \frac{n}{2} & n - 1 & a \\ \frac{1}{2} n (n - 1) & {(n - 1)}^{2} & \frac{1}{2} (n - 1) (3 n + 4) \end{matrix}|$ then the value of $\sum_{r = 1}^{n - 1} ∆_{r}$

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depends only on a

depends only on n

depends both on a and n

is independent of both a and n

answer is D.

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Detailed Solution

$∆_{r} = |\begin{matrix} r & 2 r - 1 & 3 r - 2 \\ \frac{n}{2} & n - 1 & a \\ \frac{1}{2} n (n - 1) & {(n - 1)}^{2} & \frac{1}{2} (n - 1) (3 n + 4) \end{matrix}|$
$\sum_{r = 1}^{n - 1} ∆_{r} = |\begin{matrix} \sum_{r = 1}^{n - 1} r & \sum_{r = 1}^{n - 1} (2 r - 1) & \sum_{r = 1}^{n - 1} (3 r - 2) \\ \frac{n}{2} & n - 1 & a \\ \frac{1}{2} n (n - 1) & {(n - 1)}^{2} & \frac{1}{2} (n - 1) (3 n + 4) \end{matrix}|$

$\sum_{r = 1}^{n - 1} ∆_{r} = |\begin{matrix} \frac{1}{2} n (n - 1) & {(n - 1)}^{2} & \frac{1}{2} (n - 1) (3 n + 4) \\ \frac{n}{2} & n - 1 & a \\ \frac{1}{2} n (n - 1) & {(n - 1)}^{2} & \frac{1}{2} (n - 1) (3 n + 4) \end{matrix}| = 0$