Question

If ${(1+x)}^{15}=C_0+C_1.x+C_2.x^2+....+C_{15}.x^{15}$ and $C_2+2C_3+3C_4+....+14C_{15}=a{2}^b+c,$ then find $a+b+c$

Answer 1

The required expression is $\frac{d\left(\frac{(1+x{)}^{15}}{x}\right)}{dx}{|}_{x=1}+{\frac{C_0}{x^2}}_{x=1}$
$={\frac{15x(1+x{)}^{14}-(1+x{)}^5}{x^2}}_{x=1}+1$
$={15.2}^{14}-2^{15}+1$
$=2^{14}(15-2)+1$
$={13.2}^{14}+1$
Hence $a+b+c$
$=13+14+1$
$=28$