If - r - 2 r-1 1 rr-1 sin n- x y z 2 n -1 n n-1 sin n-1 2 -sin n 2 -sin- 2- then - r-1 n - r is equal to

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Change of Variables

If Δ r = 2 ...

Question

If $Δ_{r} = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & \frac{1}{r (r + 1)} & sin n θ x & y & z 2^{n} - 1 & \frac{n}{n + 1} & \frac{sin \frac{n + 1}{2} θ sin \frac{n}{2} θ}{sin \frac{θ}{2}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$ , then $\sum_{r = 1}^{n} Δ_{r}$ is equal to

0

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Dependent

n

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Dependent of

θ

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Independent of

x, y, z

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Solution

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Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{n} & x^{n + 2} & x^{n + 4} y^{n} & y^{n + 2} & y^{n + 4} z^{n} & z^{n + 2} & z^{n + 4} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (\frac{1}{y^{2}} - \frac{1}{x^{2}}) (\frac{1}{z^{2}} - \frac{1}{y^{2}}) (\frac{1}{x^{2}} - \frac{1}{z^{2}})

- n

n > 3

\frac{1}{sin \frac{π}{n}} = \frac{1}{sin \frac{2 π}{n}} + \frac{1}{sin \frac{3 π}{n}}

{(1 + x)}^{n} = \sum_{r = 0}^{n} C_{r} x^{r}

C_{0} + \frac{C_{1}}{2} + . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

\frac{C_{0}}{x} - \frac{C_{1}}{x + 1} + \frac{C_{2}}{x + 2} - . . . . . . + {(- 1)}^{n} \frac{C_{n}}{x + n} =

C_{r}

{^{n} C}_{r}

x > m, y > n, z > r (x, y, z > 0)

∣ ∣ ∣ ∣ \begin{matrix} x & n & r m & y & r m & n & z \end{matrix} ∣ ∣ ∣ ∣ = 0

\frac{m}{x - m} - 1 + \frac{y}{y - n} - 1 + \frac{r}{z - r}

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