If $(1+x+x^2)^n = (p_0+p_1x+p_0x^2+p_3x^3+p_4x^4+\ldots++p_nx^n)$ then prove that $p_1+p4+p7 + \ldots = 3^{n-1}$

Asked Dec 27 '21 at 04:34

Active Dec 27 '21 at 04:34

Viewed 36 times

I have the following questins: If $(1+x+x^2)^n = (p_0+p_1x+p_0x^2+p_3x^3+p_4x^4+\ldots++p_nx^n)$ then prove that $p_1+p4+p7 + \ldots = 3^{n-1}$

I have taken x = 1, $\omega$, $\omega^2$ and then sum all the equations. By this, I am able to prove $p+0+p_3+p_6+\ldots=3^{n-1}$. But I cant understand how to prove $p_1+p4+p7 + \ldots = 3^{n-1}$

asked Dec 27 '21 at 04:34

Christopher Marlowe

2

$(1+x+x^2)^n$ is a polynomial of degree $\mathbf{2n}$, not $n$ – markvs Dec 27 '21 at 04:46
1

It's not $p+0$ but $p_0$. – Jean Marie Dec 27 '21 at 04:51
Are $p4 =p_4$, $p7=p_7$? – Gary Dec 27 '21 at 04:55
2

Does this answer your question? Finding sum of expressions involving coefficients of terms in the expansion $(1+x+x^2)^n$ (found using https://approach0.xyz/ formula searching tool. – Jean Marie Dec 27 '21 at 04:55

If $(1+x+x^2)^n = (p_0+p_1x+p_0x^2+p_3x^3+p_4x^4+\ldots++p_nx^n)$ then prove that $p_1+p4+p7 + \ldots = 3^{n-1}$

0 Answers0