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If \(n\) is a positive integer, then \(\sum_{r=1}^n r \cdot C_r=\)
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Verified Answer
The correct answer is:
\(n 2^{n-1}\)
\(\sum_{r=1}^n r \cdot C_r\)
\(=1 C_1+2 C_2+3 C_3+\ldots+\ldots+n C_n=n \cdot 2^{n-1}\)
\(=1 C_1+2 C_2+3 C_3+\ldots+\ldots+n C_n=n \cdot 2^{n-1}\)
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