if (1+x)15=C0+C1x+C2x2+…+C15x15 then C2+2C3+3C4+…+14C15 is equal to

# if $\left(1+\mathrm{x}{\right)}^{15}={\mathrm{C}}_{0}+{\mathrm{C}}_{1}\mathrm{x}+{\mathrm{C}}_{2}{\mathrm{x}}^{2}+\dots +{\mathrm{C}}_{15}{\mathrm{x}}^{15}$ then ${\mathrm{C}}_{2}+2{\mathrm{C}}_{3}+3{\mathrm{C}}_{4}+\dots +14{\mathrm{C}}_{15}$ is equal to

1. A

$14\cdot {2}^{14}$

2. B

$13\cdot {2}^{14}+1$

3. C

$13\cdot {2}^{14}-1$

4. D

None of these

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### Solution:

we have $\left(1+\mathrm{x}{\right)}^{15}={\mathrm{C}}_{0}+{\mathrm{C}}_{1}\mathrm{x}+{\mathrm{C}}_{2}{\mathrm{x}}^{2}+\dots +{\mathrm{C}}_{15}{\mathrm{x}}^{15}$

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

$\frac{\mathrm{x}\cdot 15\left(1+\mathrm{x}{\right)}^{14}-\left(1+\mathrm{x}{\right)}^{15}+1}{{\mathrm{x}}^{2}}={\mathrm{C}}_{2}+2{\mathrm{C}}_{3}\mathrm{x}+\dots +14{\mathrm{C}}_{15}{\mathrm{x}}^{13}$

On putting x = t, we get

${\mathrm{C}}_{2}+2{\mathrm{C}}_{3}+\dots +14{\mathrm{C}}_{15}=15\cdot {2}^{14}-{2}^{15}+1=13\cdot {2}^{14}+1$

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