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$\sum_{k=0}^{10}{ }^{20} C_k=$
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Verified Answer
The correct answer is:
$2^{19}+\frac{1}{2}{ }^{20} C_{10}$
$\sum_{K=0}^{10}{ }^{20} C_k \quad \text { i.e., }{ }^{20} C_0+{ }^{20} C_1+\ldots \ldots .+{ }^{20} C_{10}$
We know that, $(1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots .+{ }^n C_n \cdot x^n$
Put $x=1 ; 2^n={ }^n C_0+{ }^n C_1+{ }^n C_2+\ldots . .+{ }^n C_n$
Put $n=20 ; 2^{20}={ }^{20} C_0+{ }^{20} C_1+{ }^{20} C_2+\ldots \ldots .+{ }^{20} C_{20}$
$\begin{aligned} & 2^{20}+{ }^{20} C_{10}=2\left[{ }^{20} C_0+{ }^{20} C_1+\ldots \ldots+{ }^{20} C_{10}\right] \\
& {\left[{ }^{20} C_0+{ }^{20} C_1+\ldots . .+{ }^{20} C_{10}\right]=2^{19}+\frac{1}{2}{ }^{20} C_{10}} \\
& \sum_{k=0}^{10}{ }^{20} C_k=2^{19}+\frac{1}{2}{ }^{20} C_{10} \end{aligned}$
We know that, $(1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots .+{ }^n C_n \cdot x^n$
Put $x=1 ; 2^n={ }^n C_0+{ }^n C_1+{ }^n C_2+\ldots . .+{ }^n C_n$
Put $n=20 ; 2^{20}={ }^{20} C_0+{ }^{20} C_1+{ }^{20} C_2+\ldots \ldots .+{ }^{20} C_{20}$
$\begin{aligned} & 2^{20}+{ }^{20} C_{10}=2\left[{ }^{20} C_0+{ }^{20} C_1+\ldots \ldots+{ }^{20} C_{10}\right] \\
& {\left[{ }^{20} C_0+{ }^{20} C_1+\ldots . .+{ }^{20} C_{10}\right]=2^{19}+\frac{1}{2}{ }^{20} C_{10}} \\
& \sum_{k=0}^{10}{ }^{20} C_k=2^{19}+\frac{1}{2}{ }^{20} C_{10} \end{aligned}$
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