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Q.
If $(1+x)^{15}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{15} x^{15}$, then the value of $C_{2}+2 C_{3}+3 C_{4}+\ldots+14 C_{15}$ is
Binomial Theorem
Solution:
We have,$(1+x)^{15}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{15} x^{15}$
$\Rightarrow \frac{(1+x)^{15}}{x}=\frac{C_{0}}{x}+C_{1}+C_{2} x+C_{3} x^{2}+\ldots+C_{15} x^{14}$
Differentiating both sides w.r.t. $x$, we get
$\frac{x \cdot 15(1+x)^{14}-1 \cdot(1+x)^{15}}{x^{2}}$
$=-\frac{C_{0}}{x^{2}}+C_{2}+2 C_{3} x+3 C_{4} x^{2}+\ldots+14 C_{15} x^{13}$
Putting $x=1$ on both sides, we get
$15 \cdot 2^{14}-2^{15}=-C_{0}+C_{2}+2 C_{3}+3 C_{4}+\ldots+14 C_{15}$
$\Rightarrow 2^{14}(15-2)+1=C_{2}+2 C_{3}+3 C_{4}+\ldots+14 C_{15}$
$\therefore $ The given series $=2^{14} \cdot 13+1=219923$.