Use the binomial theorem to derive a closed form expression for ${n \choose 0} + {n \choose 4} + {n \choose 8} +...+{n \choose 4⌊n/4⌋}$

Question

Use the binomial theorem to derive a closed form expression for $${n \choose 0} + {n \choose 4} + {n \choose 8} +...+{n \choose 4⌊n/4⌋}$$ And I should use imaginary numbers right?

Should that last term be $\dbinom{n}{4\lfloor\tfrac{n}{4}\rfloor}$? — JimmyK4542, Sep 01 '14 at 02:45

score 5 · Accepted Answer · edited Apr 13 '17 at 12:20

5

Hint: Start with $(1+x)^n = \displaystyle\sum_{k = 0}^{n}\dbinom{n}{k}x^k$, and plug in $x = 1,i,-1,-i$. This will give you four equations. What do you get when you add these four equations together and divide by $4$?

Also see this related question.

edited Apr 13 '17 at 12:20

Community

1

answered Sep 01 '14 at 02:39

JimmyK4542

54,331

+1 Very clever solution! Can it be generalized when $4$ is replaced by another integer? EDIT: answer in the link you provided. – Kim Jong Un Sep 01 '14 at 03:02

Use the binomial theorem to derive a closed form expression for ${n \choose 0} + {n \choose 4} + {n \choose 8} +...+{n \choose 4⌊n/4⌋}$

1 Answers1