The coefficient of x 4 in the expansion of x 2- x -26 isA- 240B- -280C- -180D- 60

The correct option is C 180[x^2 (x+2)]^6 =[(x+2) x^2]^6 =∑_r=0^6^6C_r( 1)^r (x+2)^6 r(x^2)^r To get x^4 in the expansion, r can take values 0,1,2 nbsp;on ...

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Standard XII

Mathematics

General Term of Binomial Expansion

The coefficie...

Question

The coefficient of $x^{4}$ in the expansion of $(x^{2} - x - 2)^{6}$ is

60

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240

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- 280

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- 180

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Solution

The correct option is D $- 180$
${[x^{2} - (x + 2)]}^{6} = {[(x + 2) - x^{2}]}^{6} = 6 \sum r = 0^{6} C_{r} (- 1)^{r} (x + 2)^{6 - r} {(x^{2})}^{r}$

To get $x^{4}$ in the expansion, $r$ can take values $0, 1, 2$ only.
So, consider $^{6} C_{0} (x + 2)^{6} -^{6} C_{1} (x + 2)^{5} (x^{2}) +^{6} C_{2} (x + 2)^{4} {(x^{2})}^{2}$

Coefficient of $x^{4}$
$=^{6} C_{0} \cdot^{6} C_{2} \cdot 2^{2} -^{6} C_{1} \cdot^{5} C_{3} \cdot 2^{3} +^{6} C_{2} \cdot 2^{4} = 60 - 480 + 240 = - 180$

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The coefficient of x4 in the expansion of (x2−x−2)6 is

The correct option is D −180[x2−(x+2)]6=[(x+2)−x2]6=6∑r=06Cr(−1)r(x+2)6−r(x2)r To get x4 in the expansion, r can take values 0,1,2 only. So, consider 6C0(x+2)6−6C1(x+2)5(x2)+6C2(x+2)4(x2)2 Coefficient of x4 =6C0⋅6C2⋅22−6C1⋅5C3⋅23+6C2⋅24=60−480+240=−180

The coefficient of $x^{4}$ in the expansion of $(x^{2} - x - 2)^{6}$ is