Question

If $(1+x{)}^{15}=C_0+C_{1}x+C_2{x}^2+......+C_{15}{x}^{15},$ then $C_2+2C_3+3C_4+......+14C_{15}=$

• A. ${14.2}^{14}$
• B. ${13.2}^{14}+1$
• C. ${13.2}^{14}-1$
• D. None of these

Answer 1

The correct option is B ${13.2}^{14}+1$

We have $(1+x{)}^{15}=C_0+C_{1}x+C_2{x}^2+......+C_{15}{x}^{15}$
$\Rightarrow \frac{(1+x{)}^{15}-1}{x}=C_1+C_{2}x+....+C_{15}{x}^{14}$
Differentiating both sides with respect to x, we get
$=\frac{x.15(1+x{)}^{14}-(1+x{)}^{15}+1}{x^2}$
$=C_2+2C_{3}x+........{+}^{14}{C}_{15}{x}^{13}$
putting x=1, we get
$C_2+2C_3+....+14C_{15}={15.2}^{14}-2^{15}+1={13.2}^{14}+1.$