Question

Find the coefficient of $${ x }^{ 4 }$$ in the expansion of $${ \left( 2-x+{ 3x }^{ 2 } \right) }^{ 6 }$$

A
3661
B
1830
C
3300
D
none of the above

Solution

The correct option is D none of the aboveWe want the coefficient of $$x^{4}$$ in $$(2x^{0}-1x+3x^{2})^{6}$$ will be in the case $$(x^{0})^{3}(x^{1})^{1}(x^{2})^{2}$$ , $$(x^{0})^{2}(x^{1})^{4}(x^{2})^{0}$$ and $$(x^{0})^{4}(x^{1})^{0}(x^{2})^{2}$$Coefficient of $$x^{4}$$ in:$$(x^{0})^{3}(x^{1})^{1}(x^{2})^{2}=\displaystyle \frac{6!(2)^{3}(-1)^{1}(3)^{2}}{3!1!2!}=-4320$$$$(x^{0})^{2}(x^{1})^{4}(x^{2})^{0}=\displaystyle \frac{6!(2)^{2}(-1)^{4}(3)^{0}}{2!4!0!}=60$$ $$(x^{0})^{4}(x^{1})^{0}(x^{2})^{2}=\displaystyle \frac{6!(2)^{4}(-1)^{0}(3)^{2}}{4!0!2!}=2160$$ So, total will be $$2160+60-4320=-2100$$i.e., none of the above.Maths

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