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If $a_{0},a_{1},a_{2},\cdots\cdots$be coefficient in the expansion of $(1+x+x^2)^n$ in ascending power of $x$.Then prove that

$(1)\;a_{0}\cdot a_{1}-a_{1}\cdot a_{2}+a_{2}\cdot a_{3}-\cdots\cdots -a_{2n-1}\cdot a_{2n}=0$

$(2)\;a_{0}\cdot a_{2}-a_{1}\cdot a_{3}+a_{2}\cdot a_{4}-\cdots\cdots +a_{2n-2}a_{2n}=a_{n+1}$

Try: $(1+x+x^2)^n=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +a_{2n}x^{2n}$

Put $\displaystyle x=\frac{1}{x}$,we have

Try: $(1+x+x^2)^n=a_{0}x^{2n}+a_{1x^{2n+1}}+a_{2}x^{2n+2}+\cdots \cdots +a_{2n}x^{4n}$

Could someone help me to solve it? Thanks!

Jean Marie
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DXT
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1 Answers1

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Substituting $x$ by $\frac 1x$ into $$(1+x+x^2)^n=\sum_{j=0}^{2n}a_{j}x^j\tag3$$ we get $$\left(1+\frac 1x+\frac{1}{x^2}\right)^n=\sum_{j=0}^{2n}a_{j}\left(\frac 1x\right)^j\tag4$$ Multiplying the both sides by $x^{2n}$ gives $$(x^2+x+1)^n=\sum_{j=0}^{2n}a_{j}x^{2n-j}\tag5$$ Substituing $x$ by $-x$ into $(5)$ gives $$(x^2-x+1)^n=\sum_{j=0}^{2n}a_{j}(-1)^{2n-j}x^{2n-j}\tag6$$

Multiplying $(6)$ by $(3)$, we get $$(1+x^2+x^4)^{n}=\left(a_0x^{2n}-a_1x^{2n-1}+a_2x^{2n-2}-\cdots +a_{2n}\right)\left(a_0+a_1x+\cdots +a_{2n}x^{2n}\right)\tag7$$

$(1)$

Let us consider the coefficient of $x^{2n+1}$ in $(7)$.

The coefficient of $x^{2n+1}$ in the LHS of $(7)$ is $0$, so we have $$0=a_0a_1-a_1a_2+a_2a_3-\cdots -a_{2n-1}a_{2n}$$

$(2)$

Let us consider the coefficient of $x^{2n+2}$ in $(7)$.

The coefficient of $x^{2n+2}$ in the LHS of $(7)$ is the same as the coefficient of $x^{n+1}$ in $(1+x+x^2)^n$, i.e. $a_{n+1}$, so we have $$a_{n+1}=a_0a_2-a_1a_3+a_2a_4-\cdots +a_{2n-2}a_{2n}$$

mathlove
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