If (1-x+x^2)^10=a_0+a_1x+a_2x^2+ …+a_20x^20, then 2a_2+3a_3+4a_4+ …+20a_20=

Question

Accepted Answer

Correct answer: C. (c) Given expansion is (1-x+x^2)^10=a_0+a_1x+a_2x^2+ …+a_20x^20 On differentiating 10(1-x+x^2)^9 (-1+2x )=a_1+2a_2x+3a_2x^2+ …20a_20x^19 At x=1 ⇒a_1+2a_2+3a_3+ …+20a_20=10 At x=0 ⇒a_1=-10 At x=0 So, 2a_2+3a_3+ …+20a_20=20

$If (1 - x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}, then 2 a_{2} + 3 a_{3} + 4 a_{4} + \dots + 20 a_{20} =$

Solution

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