If (1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+ldots+a_{ | vedclass

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(C) Given the expansion: $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Differentiating both sides with respect to $x$:$10(1-x+x^2)^9 \cdot (

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$20$

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(C) Given the expansion: $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Differentiating both sides with respect to $x$:$10(1-x+x^2)^9 \cdot (-1+2x) = a_1 + 2a_2 x + 3a_3 x^2 + \ldots + 20a_{20} x^{19}$.Setting $x=1$:$10(1-1+1)^9 \cdot (-1+2) = a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20}$.$10(1)^9 \cdot (1) = a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20} = 10$.To find $a_1$,differentiate the original expression and set $x=0$ or observe the coefficient of $x$ in $(1-x+x^2)^{10}$:$(1-x+x^2)^{10} = 1 + 10(-x+x^2) + \ldots = 1 - 10x + \ldots$.Thus,$a_1 = -10$.Substituting $a_1 = -10$ into the equation $a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20} = 10$:$-10 + (2a_2 + 3a_3 + \ldots + 20a_{20}) = 10$.Therefore,$2a_2 + 3a_3 + \ldots + 20a_{20} = 20$.

If $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$,then $2 a_2+3 a_3+4 a_4+\ldots+20 a_{20}=$

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