If {(1 + x)^n} = {C_0} + {C₁}x + {C₂}{x^2} + .... + {Cₙ}{x^n} , then value of {C_0} + 2{C₁}

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7.Binomial Theorem

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If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$, then the value of ${C_0} + 2{C_1} + 3{C_2} + .... + (n + 1){C_n}$ will be

A

$(n + 2){2^{n - 1}}$

B

$(n + 1){2^n}$

C

$(n + 1){2^{n - 1}}$

D

$(n + 2){2^n}$

(IIT-1971)

Solution

(a) Trick: Put $n=1$, the expression is equivalent to $^1{C_0} + {2.^1}{C_1} = 1 + 2 = 3$

Only option $(a)$ gives the value.

Standard 11

Mathematics

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