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Summation by Sigma Method

If 1+ x n = C...

Question

If $1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . + C_{n} x^{n}$ then the value of $C_{0} - C_{2} + C_{4} - C_{6} + . . . . . . .$ is

A

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B

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C

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D

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Solution

The correct option is D
$S i n c e (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . + C_{n} x^{n} P u t x = i, o n b o t h t h e s i d e s, w e g e t (1 + i)^{n} = (C_{0} - C_{2} + C_{4} - . . . . .) + i (C_{1} - C_{3} + C_{5} - . . . . .) . . . . . . . . (i) A l s o, 1 + i = \sqrt{2} (c o s \frac{π}{4} + i s i n \frac{π}{4}) i n a m p l i t u d e m o d u l u s f o r m \Rightarrow (1 + i)^{n} = 2^{n / 2} {(c o s \frac{π}{4} + i s i n \frac{π}{4})}^{n} . . . . (i i) E q u a t i n g t h e r e a l p a r t s i n (i) a n d (i i) w e g e t, C_{0} - C_{2} + C_{4} - C_{6} + . . . . . = 2^{π / 2} c o s \frac{n π}{4}$

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Similar questions

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

\frac{C_{0}}{1} + \frac{C_{2}}{3} + \frac{C_{4}}{5} + . . . . =

C_{0}, C_{1}, \dots C_{n}

are the coefficient of

x

in expansion of

(1 + x)^{n}

C_{0} - C_{2} + C_{4} - C_{6} + \dots + (- 1)^{n} C_{n} =

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

, then the value of

(C_{0} - C_{1} + C_{2} - C_{3} + . . . . . + (- 1)^{n} C_{n})

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C x^{2}

C_{0} + C_{1} + C_{2} + \dots + C_{n} = 2^{n}

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

C_{0} + C_{2} + C_{4} + C_{6} + . . . . . . . . = ?

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