First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

$14 \cdot 2^{14}$
B

$13 \cdot 2^{14} + 1$
C

$13 \cdot 2^{14} - 1$
D

None of these

Solution

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

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