If $(1 + 3 x + 3 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + a_{5} x^{5} + . . . . . . + a_{40} x^{40},$ then find the value of $2 a_{2} - 6 a_{3} + 12 a_{4} - 20 a_{5} . . . . . . + 1560 a_{40}$

3450

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3350

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3540

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2150

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Solution

The correct option is D $3540$
differentiate both side twice
$20 (1 + 3 x + 3 x^{2})^{19} . (3 + 6 x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} + 5 a_{5} x^{4} + . . . . . . . . + 40 a_{40} x^{39}$
$20 [19 (1 + 3 x + 3 x^{2})^{18} (3 + 6 x)^{2} + (1 + 3 x + 3 x^{2})^{19} (6)] = 2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + 20 a_{5} x^{3} + . . . . . + 1560 a_{40} x^{38}$
Now substitute $x = - 1$ in both sides
$20 [19.1 (- 3)^{2} + 1.6] = 2 a_{2} - 6 a_{3} + 12 a_{4} - 20 a_{5} + . . . . . + 1560 a_{40}$
or $2 a_{2} - 6 a_{3} + 12 a_{4} - 20 a_{5} + . . . . . . + 1560 a_{40} = 3540$

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