Given (1 - 2x + 5x^2 - 10x^3) (1 + x)^n = 1 + a₁x + a₂x^2 + .... and that a₁^2,= 2a₂ then va

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7.Binomial Theorem

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Given $(1 - 2x + 5x^2 - 10x^3) (1 + x)^n = 1 + a_1x + a_2x^2 + ....$ and that $a_1^2\,= 2a_2$ then the value of $n$ is

A

$6$

B

$2$

C

$5$

D

$3$

Solution

$( 1 – 2x + 5x^2 + 10x^3)$ $[ C_0 + C_1x + C_2x^2 + ….] $ $= 1 + a_1x + a_2x^2 + …..$

$a_1 = n – 2$ and $a_2 = \frac{{n(n – 1)}}{2}\,\, – \,\,2n\,\, + \,5$

put $a_1^2\,= 2a_2$

$(n – 2)^2 = n (n – 1) – 4n + 10$

$n^2 – 4n + 4 = n^2 – 5n + 10$

$n = 6$

Standard 11

Mathematics

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