{C_0} - {C₁} + {C₂} - {C₃} + ..... + {( - 1)^n}{Cₙ} is equal to

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7.Binomial Theorem

easy

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

A

${2^n}$

B

${2^n} - 1$

C

$0$

D

${2^{n - 1}}$

Solution

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

Standard 11

Mathematics

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