Question

If $(1+x{)}^{15}=C_0+C_{1}x+C_2{x}^2+\dots +C_{15}{x}^{15}$, then the value of $C_2+2C_3+3C_4+\dots +14C_{15}$ is

• A. 219923
• B. 16789
• C. 219982
• D. none of these

Answer 1

Answer: (A)

We have,$(1+x{)}^{15}=C_0+C_{1}x+C_2{x}^2+\dots +C_{15}{x}^{15}$
$\Rightarrow \frac{(1+x{)}^{15}}{x}=\frac{C_0}{x}+C_1+C_{2}x+C_3{x}^2+\dots +C_{15}{x}^{14}$
Differentiating both sides w.r.t. $x$, we get
$\frac{x\cdot 15(1+x{)}^{14}-1\cdot (1+x{)}^{15}}{x^2}$
$=-\frac{C_0}{x^2}+C_2+2C_{3}x+3C_4{x}^2+\dots +14C_{15}{x}^{13}$
Putting $x=1$ on both sides, we get
$15\cdot 2^{14}-2^{15}=-C_0+C_2+2C_3+3C_4+\dots +14C_{15}$
$\Rightarrow 2^{14}(15-2)+1=C_2+2C_3+3C_4+\dots +14C_{15}$
$\therefore$ The given series $=2^{14}\cdot 13+1=219923$.