If(1+x)n=C0+C1x+C2x2+…+Cnxn them the value of C0+2C1+3C2+…+(n+1)Cn will be

# If$\left(1+\mathrm{x}{\right)}^{\mathrm{n}}={\mathrm{C}}_{0}+{\mathrm{C}}_{1}\mathrm{x}+{\mathrm{C}}_{2}{\mathrm{x}}^{2}+\dots +{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ them the value of ${\mathrm{C}}_{0}+2{\mathrm{C}}_{1}+3{\mathrm{C}}_{2}+\dots +\left(\mathrm{n}+1\right){\mathrm{C}}_{\mathrm{n}}$ will be

1. A

$\left(\mathrm{n}+2\right){2}^{\mathrm{n}-1}$

2. B

$\left(\mathrm{n}+1\right){2}^{\mathrm{n}}$

3. C

$\left(\mathrm{n}+1\right){2}^{\mathrm{n}-1}$

4. D

$\left(\mathrm{n}+2\right){2}^{\mathrm{n}}$

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### Solution:

Since, $\mathrm{x}\left(1+\mathrm{x}{\right)}^{\mathrm{n}}={\mathrm{xC}}_{0}+{\mathrm{C}}_{1}{\mathrm{x}}^{2}+{\mathrm{C}}_{2}{\mathrm{x}}^{3}+\dots +{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1}$

On differentiating w.r.t. x, we get

$\left(1+\mathrm{x}{\right)}^{\mathrm{n}}+\mathrm{nx}\left(1+\mathrm{x}{\right)}^{\mathrm{n}-1}={\mathrm{C}}_{0}+2{\mathrm{C}}_{1}\mathrm{x}+3{\mathrm{C}}_{2}{\mathrm{x}}^{2}$$+\dots +\left(\mathrm{n}+1\right){\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$

Put x = L, we get ${\mathrm{C}}_{0}+2{\mathrm{C}}_{1}+3{\mathrm{C}}_{2}+\dots +\left(\mathrm{n}+1\right){\mathrm{C}}_{\mathrm{n}}$

$={2}^{\mathrm{n}}+\mathrm{n}{2}^{\mathrm{n}-1}={2}^{\mathrm{n}-1}\left(\mathrm{n}+2\right)$

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