Question

Passage:If n is a positive and $$a_{1},\ a_{2},\ a_{3},..a_{m}\in C$$ then$$(a_{1}+a_{2}+a_{3}+.... +a_{m})^{n}=\displaystyle \Sigma(\frac{n!}{{ n }_{ 1 }!{ n }_{ 2 }!{ n }_{ 3 }!\ldots n_{m}!})a_{1}^{{ n }_{ 1 }}a_{2}^{{ n }_{ 2 }}a_{3}^{{ n }_{ 3 }}....a_{m}^{{ n }_{ m }}$$where $$n_{1},n_{2},\ n_{3},\ n_{m}$$ are all non negative integers subject to the condition $$n_{1}+n_{2}+n_{3}+\ldots+n_{m}=n$$The coefficient of $$x^{39}$$ in the expansion of $$(1+x+2x^{2})^{20}$$ is

A
5×219
B
5×230
C
5×221
D
5×223

Solution

The correct option is D $$5\times 2^{21}$$$$(1+x+2x^2)^{20}$$$$x^{39}$$ can be written as $$1.x.(2x^2)^{19}$$Hence coefficient will be $$\dfrac{20!}{(20-19)1!(20-19)1!(20-1)!}.2^{19}$$$$=\dfrac{20!}{1!1!19!}.2^{19}$$$$=20.2^{19}$$$$=4.5.2^{19}$$$$=5.2^{21}$$Maths

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