Let $S$ $=$ $\binom{n}3 $ $+$ $\binom{n}7$ $+$ $\binom{n}{11} $ $.....$
Find $S$.
This question can be solved by taking the difference of the binomial series of $\frac{-(1+x)^n - (1-x)^n}{2}$ and $\frac{(1+ix)^n - (1-ix)^n}{2i}$ , where $i^2$ $=$ $-1$ and substituting $x$ $=$ $1$ and dividing the resulting value by 2.
However, I wonder if the value of $S$ can be evaluated using calculus ? Please help me with this thought.
Let $$S=\sum_{k=0}^{n} {n \choose 4k+3}$$ $$(1+x)^n=1+{n \choose 1} x+ {n \choose 2} x^2+ {n \choose 3}x^3+ {n \choose 4} x^4+...+{n \choose n}x^n~~~(1)$$ Let $z^4=1$ its roots are $1,a,a^2,a^3 (1,i,-1,-i)$, with $1+a+a^2+a^3=0$ Put $x=w$ in (1), then putting $x=1,a,a^2,a^3(1,i,-1,-i)$ in (1) we get $$2^n=\sum_{k=0}^{n} {n \choose k},~ (1+a)^n=\sum_{k=0}^{n} {n \choose k} a^k, ~(1+a^2)^n=\sum_{k=0}^n {n \choose k} a^{2k},$$ $$~(1+a^3)^n=\sum_{k=0}^{n} {n \choose k} a^{3k}$$ Multiplying these identities by $1, a, a^2, a^3$ respectively and adding them we get $$2^n+a(1+a)^n+a^2(1+a^2)^n+a^3(1+a^3)= \sum_{k=0}^n (1+a^{k+1}+a^{2k+2}+a^{3k+3}) {n \choose k}~~~(2)$$ The term in the parenthesis in RHS of (2) is 1 only for$ k=3,7,11,15,...$, otherwise it vanishes. So $$S=2^n+(1+i)^n+i^2(1+i^2)^n+i^3(1+i^3)^n= 2^ni[(1+i)^n-(1-i)^n]$$ $$=2^n-2^{n/2}~ 2\sin n\pi/4$$