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If $\mathrm{c}_0, \mathrm{c}_1, \mathrm{c}_2, \ldots \ldots \ldots \ldots \ldots . \mathrm{c}_{\mathrm{n}}$ denote the co-efficients in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$ then the value of $\mathrm{c}_1+2 \mathrm{c}_2+3 \mathrm{c}_3+\ldots . .+\mathrm{nc}_{\mathrm{n}}$ is
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Verified Answer
The correct answer is:
$\mathrm{n} \cdot 2^{\mathrm{n}-1}$
Hints : $(1+\mathrm{x})^{\mathrm{n}}=\dot{c}_0+\mathrm{xc}_1+\mathrm{x}^2 \mathrm{c}_2+\ldots \ldots . \mathrm{x}^{\mathrm{n}} \mathrm{c}_{\mathrm{n}}$
$n(1+x)^{n-1}=c_1+2 x c_2+\ldots \ldots . . n x^{n-1} c_n$
Put $x=1$
$$
\mathrm{n}(2)^{\mathrm{n}-1}=\mathrm{c}_1+2 \mathrm{c}_2+3 \mathrm{c}_2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{nc}_{\mathrm{n}}
$$
$n(1+x)^{n-1}=c_1+2 x c_2+\ldots \ldots . . n x^{n-1} c_n$
Put $x=1$
$$
\mathrm{n}(2)^{\mathrm{n}-1}=\mathrm{c}_1+2 \mathrm{c}_2+3 \mathrm{c}_2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{nc}_{\mathrm{n}}
$$
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