How is $ C_0^n+\frac{1}{2}(C_0^n-C_1^n)+\frac{1}{3}(C_0^n-C_1^n+C_2^n)+.. =C_0^{n-1}+\frac{1}{2}C_1^{n-1}(-1)+\frac{1}{3}C_2^{n-1}(-1)^2... $?

Question

I saw this result in the solution of a question: $ C_0^n + \frac{1}{2}(C_0^n-C_1^n) + \frac{1}{3}(C_0^n-C_1^n+C_2^n)+... = C_0^{n-1} + \frac{1}{2}C_1^{n-1}(-1) + \frac{1}{3}C_2^{n-1}(-1)^2... = 1/n $. How is this result obtained? I have to solve this from the RHS since this is a step of a bigger question. How do I approach this?

Answer 1

By the binomial theorem, $$(1+z(-1))^{n-1}=C_0^{n-1}+C_1^{n-1}z(-1)+C_2^{n-1}z^{2}(-1)^2+\dots$$ Taking the definite integral with respect to $z$ of both sides from $0$ to $x$, $$\int_0^x(1+z(-1))^{n-1}dz=\int_0^xC_0^{n-1}+C_1^{n-1}z(-1)+C_2^{n-1}z^{2}(-1)^2+\dots dz$$ and therefore $$-\frac{(1+x(-1))^n}{n}+\frac{1}{n}=C_0^{n-1}x+C_1^{n-2}\frac{x^2}{2}(-1)+C_2^{n-1}\frac{x^{3}}{3}(-1)^2+\dots$$ Evaluating at $x=1$ gives the result.