If (1 + x)^{15} = C_0 + C_1x + C_2x^2 + ..... | vedclass

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(B) Given $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + .... + C_{15}x^{15}$.Subtracting $C_0 = 1$ and dividing by $x$,we get:$\frac{(1 + x)^{15} - 1}{x} = C_

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

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(B) Given $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + .... + C_{15}x^{15}$.Subtracting $C_0 = 1$ and dividing by $x$,we get:$\frac{(1 + x)^{15} - 1}{x} = C_1 + C_2x + C_3x^2 + .... + C_{15}x^{14}$.Differentiating both sides with respect to $x$:$\frac{d}{dx} \left( \frac{(1 + x)^{15} - 1}{x} \right) = C_2 + 2C_3x + 3C_4x^2 + .... + 14C_{15}x^{13}$.Using the quotient rule on the left side:$\frac{x \cdot 15(1 + x)^{14} - ((1 + x)^{15} - 1)}{x^2} = C_2 + 2C_3x + 3C_4x^2 + .... + 14C_{15}x^{13}$.Putting $x = 1$:$C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = \frac{1 \cdot 15(2)^{14} - (2^{15} - 1)}{1^2}$.$= 15 \cdot 2^{14} - 2^{15} + 1$.$= 15 \cdot 2^{14} - 2 \cdot 2^{14} + 1$.$= (15 - 2) \cdot 2^{14} + 1 = 13 \cdot 2^{14} + 1$.

If $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + ...... + C_{15}x^{15},$ then $C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = $

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