If $(1 + a x + b x^{2} + c x^{3})^{10} = 1 + p_{1} x + p_{2} x^{2} + . . . . . p_{30} x^{30}, p_{1} = 200, p_{2} = 200, p_{3} = 1000$ . Find the value of b.

- 1780

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1780

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1820

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Solution

The correct option is A $- 1780$
$(1 + a x + b x^{2} + c x^{3})^{10} = 1 + b_{1} x + b_{2} x^{2} + . . . . . + b_{30} 2^{30}$
$b_{1} = 200; b_{2} = 200$
Let $y = a x + b x^{2} + c x^{3}$
$(1 + y)^{10} =^{10} C_{0} +^{10} C_{1} +^{10} C_{2} y^{2} + . . . . . .$
$= 1 + 10 (a x + b x^{2} + c x^{2}) + 45 (a x + b x + c x^{2})^{2}$
$= 1 + 10 a x + 10 b x^{2} + 10 c x^{2} + 45 a^{2} + . . . . . .$
$∴ 10 a x = b_{1} x$
$10 a = 200$
$a = 20$
$(10 b + 45 a^{2}) x^{2} = b 2 x^{2}$
$10 b + 18000 = 200$
$10 b = - 17800$
$b = - 1780$

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