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If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$, then the value of $C_0+C_2+C_4+C_6+\ldots .$. is
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Verified Answer
The correct answer is:
$2^{n-1}$
$(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots .+C_n x^n$
Putting $x=1$, we get
$\Rightarrow 2^n=C_0+C_1+C_2+\ldots . .+C_n...(i)$
or
$C_1+C_2+C_3+\ldots .+C_n=2^n-1 \quad\left[\boxtimes C_0={ }^n C_0=1\right]$
Again, putting $x=-1$, we get
$0=C_0-C_1+C_2-C_3+\ldots$
or
$C_0+C_2+C_4+\ldots=C_1+C_3+C_5+\ldots \text { i.e. } A=B$
Also from (i), $A+B=2^n$ or $A=2^{n-1}=B$
Hence, $C_0+C_2+C_4+\ldots=2^{n-1}$
Putting $x=1$, we get
$\Rightarrow 2^n=C_0+C_1+C_2+\ldots . .+C_n...(i)$
or
$C_1+C_2+C_3+\ldots .+C_n=2^n-1 \quad\left[\boxtimes C_0={ }^n C_0=1\right]$
Again, putting $x=-1$, we get
$0=C_0-C_1+C_2-C_3+\ldots$
or
$C_0+C_2+C_4+\ldots=C_1+C_3+C_5+\ldots \text { i.e. } A=B$
Also from (i), $A+B=2^n$ or $A=2^{n-1}=B$
Hence, $C_0+C_2+C_4+\ldots=2^{n-1}$
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