C_0 - C_1 + C_2 - C_3 + ..... + ( - 1)^n C_n is equal to - Vidyadip

MCQ

${C_0} - {C_1} + {C_2} - {C_3} + ..... + {( - 1)^n}{C_n}$ is equal to

A
${2^n}$
B
${2^n} - 1$
✓
$0$
D
${2^{n - 1}}$

✓

Answer

Correct option: C.

$0$

c
(c) We know that${(1 + x)^n} = {\,^n}{C_0} + {\,^n}{C_1}x + {\,^n}{C_2}{x^2} + .... + {\,^n}{C_n}{x^n}$

Putting $x = -1$, we get ${(1 - 1)^n} = {\,^n}{C_0} - {\,^n}{C_1} + {\,^n}{C_2} - .....{( - 1)^{n\,\,n}}{C_n}$

Therefore ${C_0} - {C_1} + {C_{_2}} - {C_3} + ....( - 1){\,^n}{C_n} = 0$

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