Online Course Discussion Forum

Math Challenge III Problems 2.33 and 2.34

 
 
Picture of James Shi
Math Challenge III Problems 2.33 and 2.34
by James Shi - Friday, 24 July 2020, 7:46 PM
 

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.

 
Picture of Dr. Kevin Wang
Re: Math Challenge III Problems 2.33 and 2.34
by Dr. Kevin Wang - Friday, 24 July 2020, 10:00 PM
 

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 $x(1+x)^n$?