If Dp - p 15 8 p2 55 9 p3 225 10 - then D1-D2-D3-D4-D5 is equal to

The correct option is A 0 D_1+D_2+D_3+D_4+D_5 =1 amp; #160; 15 amp; #10; 8 #160; 1 amp;55 #160; amp;9 #160; 1 amp; 22 ...

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Properties of Determinants

if Dp = p...

Question

if $D_{p}$ = $∣ ∣ ∣ ∣ ∣ \begin{matrix} p & 15 & 8 p^{2} & 55 & 9 p^{3} & 225 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then $D_{1} + D_{2} + D_{3} + D_{4} + D_{5}$ is equal to

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625

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125

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Solution

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Similar questions

D_{P} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} P & 15 & 8 P^{2} & 35 & 9 P^{3} & 25 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣,

D_{1} + D_{2} + D_{3} + D_{4} + D_{5}

D_{p} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} p & 15 & 8 p^{2} & 35 & 9 p^{3} & 25 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣

D_{1} + D_{2} + D_{3} + D_{4} + D_{5}

1, d_{1}, d_{2}, d_{3}, d_{4}

x^{5} = 1

E = \frac{ω - d_{1}}{ω^{2} - d_{1}} \cdot \frac{ω - d_{2}}{ω^{2} - d_{2}} \cdot \frac{ω - d_{3}}{ω^{2} - d_{3}} \cdot \frac{ω - d_{4}}{ω^{2} - d_{4}}

ω

d_{1}, d_{2}, d_{3}

d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}

\frac{a △^{2}}{r^{2}}

a

(- 1, 2, 3)

d_{1}, d_{2}, d_{3}

d_{4}, d_{5}, d_{6}

\sum_{i = 1}^{6} d_{i}

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