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The sum of all Coefficient of even power of $x$ in

$(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n}),n\in \mathbb{N}$

what i try

for $n=1,$ we have

$(1-x+x^2)(1+x+x^2)=(1+1)-(1)+(1+1)=3$

for $n=2,$ we have

$(1-x+x^2-x^3+x^4)(1+x+x^2+x^3+x^4)$

$=(1+1+1)-(1+1)+(1+1+1)-(1+1)+(1+1+1)=5$

so in this way , get sum of coefficient in original expression is $2n+1$

but How do i solve it without substituting value of $n$, Help me

Community
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jacky
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3 Answers3

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These are two standard geometric series, one with ratio $-x$, the other with ratio $x$. Their product is

$(1-(-x)^{2n+1})(1-x^{2n+1})/((1+x)(1-x))\\ =(1+x^{2n+1})(1-x^{2n+1})/(1-x^2)\\ =(1-x^{4n+2})/(1-x^2) $

At $x=1$, applying Hoppy, we get

$(4n+2)x^{4n+1}/(2x) =(2n+1)x^{4n} \to 2n+1 $.

marty cohen
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  • Thanks Martin, Why we apply $\lim_{x\rightarrow 1}$ – jacky Jan 07 '20 at 10:16
  • Thanks Marty Cohen got it. $\displaystyle \frac{1-(x^2)^{2n+1}}{1-x^2}=\frac{(1-x^2)(1+x^2+x^4+x^6+\cdots+x^{2n})}{1+x^2}$. So we have $=1+x^2+x^4+\cdots+x^{2n}$ – jacky Jan 07 '20 at 14:11
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$$f(x)=(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n})$$ $$f(x)=\Bigr[(1+x^2+\cdots +x^{2n})-(x+x^3+\cdots +x^{2n-1})\Bigr]\times \Bigr[(1+x^2+\cdots +x^{2n})+(x+x^3+\cdots +x^{2n-1})\Bigr]$$ $$f(x)=(1+x^2+\cdots +x^{2n})^2-(x+x^3+\cdots +x^{2n-1})^2$$ $$f(x)=\biggr(\frac{1-x^{2n}}{1-x^2}\biggr)^2-\biggr(\frac{x-x^{2n-1}}{1-x^2}\biggr)^2$$ $$f(x)=\frac{1+x^{4n}-2x^{2n}-x^2-x^{4n-2}+2x^{2n}}{(1-x^2)^2}$$ $$f(x)=\frac{1-x^2+x^{4n}-x^{4n-2}}{(1-x^2)^2}$$ $$f(x)=\frac{(1-x^2)(1-x^{4n-2})}{(1-x^2)^2}$$ $$f(x)=\frac{1-x^{4n-2}}{1-x^2}$$

Sam
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Let's call sum of odd coefficients $C_O$ and sum of even coefficients $C_E$. You can write your polynomial as $$P(x)=O(x)+E(x)$$, where $O(x)$ is the sum of odd powered terms and $E(x)$ is the sum of even powered terms. Note that if you plug in $x=1$, you get $$C_O=O(x=1)$$ and $$C_E=E(x=1)$$ Similarly, if you plug in $x=-1$, the odd terms will acquire a minus sign, while the even ones will be unchanged. So $$O(x=-1)=-C_O$$ and $$E(x=-1)=C_E$$ Then $$C_E=\frac12 (E(1)+E(-1))=\frac12(E(1)+E(-1))+\frac12(O(1)+O(-1))=\frac12(P(1)+P(-1))$$ If you plug in $x=1$ in your original form, the second parenthesis will evaluate to $2n+1$, and the first parenthesis will be $1$. If you put in $x=-1$, you have the first parenthesis $2n+1$ and the second one $1$. Then $$C_E=2n+1$$

Andrei
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