if $(1+\mathrm{x}{)}^{15}={\mathrm{C}}_0+{\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2{\mathrm{x}}^2+\dots +{\mathrm{C}}_{15}{\mathrm{x}}^{15}$ then ${\mathrm{C}}_2+2{\mathrm{C}}_3+3{\mathrm{C}}_4+\dots +14{\mathrm{C}}_{15}$ is equal to 

Solution:

we have $(1+\mathrm{x}{)}^{15}={\mathrm{C}}_0+{\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2{\mathrm{x}}^2+\dots +{\mathrm{C}}_{15}{\mathrm{x}}^{15}$

$\Rightarrow \frac{(1+\mathrm{x}{)}^{15}-1}{\mathrm{x}}={\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{x}+\dots +{\mathrm{C}}_{15}{\mathrm{x}}^{14}$

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

$\frac{\mathrm{x}\cdot 15(1+\mathrm{x}{)}^{14}-(1+\mathrm{x}{)}^{15}+1}{{\mathrm{x}}^2}={\mathrm{C}}_2+2{\mathrm{C}}_3\mathrm{x}+\dots +14{\mathrm{C}}_{15}{\mathrm{x}}^{13}$

On putting x = t, we get 

${\mathrm{C}}_2+2{\mathrm{C}}_3+\dots +14{\mathrm{C}}_{15}=15\cdot 2^{14}-2^{15}+1=13\cdot 2^{14}+1$