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2 years ago

3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20n%5C%5C%20evaluate%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%5C%3A%20%CE%A3%20%20%5C%3A%20%28nCi%29%5C%5C%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20i%20%3D%200" id="TexFormula1" title=" \: \: \: \: \: \: \: \: \: \: \: n\\ evaluate \: \: \: \: \: Σ \: (nCi)\\ \: \: \: \: \: \: \: \: \: \: \: \: i = 0" alt=" \: \: \: \: \: \: \: \: \: \: \: n\\ evaluate \: \: \: \: \: Σ \: (nCi)\\ \: \: \: \: \: \: \: \: \: \: \: \: i = 0" align="absmiddle" class="latex-formula">
Evaluate the summation

Mathematics

1 answer:

Sedaia [141]2 years ago

7 0

Assuming you mean

$\displaystyle \sum_{i=0}^n {}_nC_{i}$

where

${}_n C_i = \dbinom ni = \dfrac{n!}{i! (n-i)!}$

we have by the binomial theorem

$\displaystyle (1 + 1)^n = \sum_{i=0}^n {}_nC_{i} \cdot 1^i \cdot 1^{n-i}$

so that the given sum has a value of $\boxed{2^n}$ .

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