Binomial Coefficient Equivalency

TranscendArcu
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Find an expression that is identical to \sum_{k=0}^n \binom{3n}{3k}

According to Wolfram, the correct solution to this is: \frac{1}{3} \left(2(-1)^n + 8^n\right)

But I'm not sure which identities of the binomial coefficient I'm supposed to use to prove this. Can anyone give me some direction?

Thanks!
 
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Does nobody have any ideas? I was wondering if it were possible to confirm Wolfram's answer via induction, but expanding the resulting binomial coefficients fron the n-1 to the n case is proving to be fairly difficult. Any help is appreciated.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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