If c_0, c_1, c_2, ldots, c_n denote the coeff | vedclass

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(A) The binomial expansion is given by: $(1+x)^n = c_0 + c_1x + c_2x^2 + \ldots + c_nx^n$. Differentiating both sides with respect to $x$, we get: $n(

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$n \cdot 2^{n-1}$

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(A) The binomial expansion is given by: $(1+x)^n = c_0 + c_1x + c_2x^2 + \ldots + c_nx^n$. Differentiating both sides with respect to $x$, we get: $n(1+x)^{n-1} = c_1 + 2c_2x + 3c_3x^2 + \ldots + nc_nx^{n-1}$. To find the sum $c_1 + 2c_2 + 3c_3 + \ldots + nc_n$, we substitute $x = 1$ into the differentiated equation: $n(1+1)^{n-1} = c_1 + 2c_2(1) + 3c_3(1)^2 + \ldots + nc_n(1)^{n-1}$. This simplifies to: $n(2)^{n-1} = c_1 + 2c_2 + 3c_3 + \ldots + nc_n$. Thus, the value is $n \cdot 2^{n-1}$.

If $c_0, c_1, c_2, \ldots, c_n$ denote the coefficients in the expansion of $(1+x)^n$, then the value of $c_1 + 2c_2 + 3c_3 + \ldots + nc_n$ is

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