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If $\mathrm{C}_{0}, \mathrm{C}_{1}, \mathrm{C}_{2}, \ldots \ldots \ldots \ldots . \mathrm{C}_{\mathrm{n}}$ denote the binomial
coefficientsin the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, then the value of $\mathrm{C}_{0}+\left(\mathrm{C}_{0}+\mathrm{C}_{1}\right)+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\mathrm{C}_{2}\right)+\ldots$
$+\left(C_{0}+C_{1}+\ldots . .+C_{n-1}\right)$
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coefficientsin the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, then the value of $\mathrm{C}_{0}+\left(\mathrm{C}_{0}+\mathrm{C}_{1}\right)+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\mathrm{C}_{2}\right)+\ldots$
$+\left(C_{0}+C_{1}+\ldots . .+C_{n-1}\right)$
Solution:
2991 Upvotes
Verified Answer
The correct answer is:
$\mathrm{n} \cdot 2^{\mathrm{n}-1}$
$\begin{aligned} & \mathrm{C}_{0}+\left(\mathrm{C}_{0}+\mathrm{C}_{1}\right)+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\mathrm{C}_{2}\right)+\ldots \ldots \ldots \ldots \ldots \\ &+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\ldots \ldots \ldots . \mathrm{C}_{\mathrm{n}-1}\right) \\ &=\mathrm{n} \mathrm{C}_{0}+(\mathrm{n}-1) \mathrm{C}_{1}+(\mathrm{n}-2) \mathrm{C}_{2}+\ldots \ldots \mathrm{C}_{\mathrm{n}-1} \\ &=\mathrm{C}_{1}+2 \mathrm{C}_{2}+3 \mathrm{C}_{3}+4 \mathrm{C}_{4} \ldots \ldots \mathrm{n} \mathrm{C}_{\mathrm{n}}=\mathrm{n} .2^{\mathrm{n}-1} \end{aligned}$
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