If cos - - i sin - is a root of the equation xn - a1 xn-1- a2xn-2 - - an-1x-an - 0 then the value of - r -1n ar cos r - equals where all coefficient are real

The correct option is B 1 #160; 1+x^4+a_1x^x 1+a_2x^x 1+...+a_n = 1 ∑_r=1^n a_r cos r θ #160; #160; ⋆ A/C to given eq^n we can ...

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Order of a Differential Equation

If cos θ + ...

Question

If $c o s θ + i s i n θ$ is a root of the equation $x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . . . . . . . + a_{n - 1} x + a_{n} = 0$ then the value of $\sum_{r = 1}^{n} a_{r} c o s r θ$ equals (where all coefficient are real)

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cos θ + i sin θ

x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 1} + . . . . . . . + a_{n - 1} x + a_{n} = 0

n \sum r = 1 a_{r} cos r θ =

a_{1}, a_{2}, a_{3}, \dots, a_{n}; (n \geq 2)

(n - 1) a_{1}^{2} - 2 n a_{2} < 0,

x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0

\frac{d}{d x} [x^{n} - a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . + {(- 1)}^{n} a_{n}] e^{x} = x^{n} e^{x}

\frac{n!}{r!}

\frac{(n - r)!}{r!}

\frac{n!}{(n - r)!}

z = cos θ + i sin θ

a_{0} z^{n} + a_{1} z^{n - 1} + a_{2} z^{n - 2} + \dots \dots + a_{n - 1} z + a_{n} = 0

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