Let Δa=|(a−1)n6(a−1)22n24n−2(a−1)33n33n2−3n| the value of ∑a=1nΔa is

∑a=1nΔa=(n−1)n2n6(n−1)n(2n−1)62n24n−2(n−1)2n243n33n2−3n =(n−1)n221162n−132n4n−2(n−1)n23n23n2−3n =(n−1)n221102n−132n0(n−1)n23n20=0

Let Δa=|(a−1)n6(a−1)22n24n−2(a−1)33n33n2−3n| the value of ∑a=1nΔa is

∑a=1nΔa=(n−1)n2n6(n−1)n(2n−1)62n24n−2(n−1)2n243n33n2−3n =(n−1)n221162n−132n4n−2(n−1)n23n23n2−3n =(n−1)n221102n−132n0(n−1)n23n20=0

AI Mentor

Check Your IQ

Free Expert Demo

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 $Δ_{a} = | \begin{matrix} (a - 1) & n & 6 \\ {(a - 1)}^{2} & 2 n^{2} & 4 n - 2 \\ {(a - 1)}^{3} & 3 n^{3} & 3 n^{2} - 3 n \end{matrix} |$ the value of $\sum_{a = 1}^{n} Δ_{a}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$0$

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

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

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

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\sum_{a = 1}^{n} Δ_{a} = |\begin{matrix} \frac{(n - 1) n}{2} & n & 6 \\ \frac{(n - 1) n (2 n - 1)}{6} & 2 n^{2} & 4 n - 2 \\ \frac{{(n - 1)}^{2} n^{2}}{4} & 3 n^{3} & 3 n^{2} - 3 n \end{matrix}| = \frac{(n - 1) n^{2}}{2} |\begin{matrix} 1 & 1 & 6 \\ \frac{2 n - 1}{3} & 2 n & 4 n - 2 \\ \frac{(n - 1) n}{2} & 3 n^{2} & 3 n^{2} - 3 n \end{matrix}| = \frac{(n - 1) n^{2}}{2} |\begin{matrix} 1 & 1 & 0 \\ \frac{2 n - 1}{3} & 2 n & 0 \\ \frac{(n - 1) n}{2} & 3 n^{2} & 0 \end{matrix}| = 0$