if (1+x)15=C0+C1x+C2x2+…+C15x15 then C2+2C3+3C4+…+14C15 is equal to - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

we have $(1 + x)^{15} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{15} x^{15}$

$\Rightarrow \frac{(1 + x)^{15} - 1}{x} = C_{1} + C_{2} x + \dots + C_{15} x^{14}$

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

$\frac{x \cdot 15 (1 + x)^{14} - (1 + x)^{15} + 1}{x^{2}} = C_{2} + 2 C_{3} x + \dots + 14 C_{15} x^{13}$

On putting x = t, we get

$C_{2} + 2 C_{3} + \dots + 14 C_{15} = 15 \cdot 2^{14} - 2^{15} + 1 = 13 \cdot 2^{14} + 1$

if $(1 + x)^{15} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{15} x^{15}$ then $C_{2} + 2 C_{3} + 3 C_{4} + \dots + 14 C_{15}$ is equal to