Online Course Discussion Forum
Math Challenge III Problems 2.33 and 2.34
These two problems are very similar. I know how to find the sum of the answers to the two problems since $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \dots + (-1)^{n}\binom{n}{n} = 0$. If I add the result from example 2.14 to this identity, I get the sum of the two answers. However, I do not know how to find the separate answers. I know it is related to the binomial theorem since I read the solution to example 2.14. How should I approach this problem? Thank you.
The problem is to calculate the sum \[\binom{n}{1} + \binom{n}{4}+\binom{n}{7}+\cdots \]
and similar sum. Considering problem 2.14, if you understand the solution, then you see it is the same idea as the proof of the identity you mentioned. We use the fact that $\omega^2 + \omega + 1=0$ and $\omega^3=1$. For the binomial expansion, instead of expanding $(1+x)^n$, what if you expand $x(1+x)^n$?
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