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I want to find the solutions of $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$.

It is not hard to see $x=0$ is a root of the equation. but I don't know how to solve this equation in general. I can see terms of the equation looks very similar to binomial expansion of $[(x+1)+(x-1)]^{63}$ except the coefficient of each term is $1$ rather than $63\choose k $ (for $k=0,1,\cdots,63$ ). is it possible to use binomial theorem to solve the equation? (or other approaches)

Etemon
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3 Answers3

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Note that $x=-1$ is not a solution. This is a GP with common ratio $r=(x-1)/(x+1)\ne1$ and the sum is$$(x+1)^{63}\left[\frac{r^{64}-1}{r-1}\right]=0\implies r=\pm1$$Since $r\ne1$ it must equal $-1$ giving only real solution as $x=0$.

Shubham Johri
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$$\frac {a^n - b^n}{a-b} = a^{n-1} + a^{n-2} b + a^{n-3}b^2 + \dots + ab^{n-2} + b^{n-1}$$

Hence:

$$(x+1)^{63} + (x+1)^{62}(x-1) + \dots+ (x-1)^{63} = \frac {(x+1)^{64}-(x-1)^{64}}{(x+1)-(x-1)} = \frac12((x+1)^{64} - (x-1)^{64})$$

If the above expression is zero, we must have $(x+1)^{64} = (x-1)^{64}$. Hence $\dfrac {x+1}{x-1}$ must be one of the 64-th roots of unity (or if you are only considering real roots, $\pm 1$).

player3236
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You can solve it by using the formula of GP
Here $a = (x+1)^{63}$ and $r = \dfrac{x-1}{x+1}$

19aksh
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Shrivardhan
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  • See Shubham's solution how to sum a finite number of terms GP. So yes its a GP, but you dont say how to sum it to a specific number of terms - and if you could, you have not said how that finds the solutions. – Stilez Feb 09 '21 at 11:34