I have the set $A=\left\{1+\displaystyle\sum_{i=1}^n (3-(-1)^i)\;\text{ where }\;n\in\mathbb{N}_0\right\}$ and I have to prove equality with $B=\{x\in\mathbb{N}\;\text{ where }\;2 \text{ and } 3 \text{ are not factors of }x\}$ and my first thought was to transform the sum to get a function which returns the numerical value $f(n)\in\mathbb{Z}$ based on the parameter $n$. But I don't know which methods i can use to "transform" or "expand" this sum (what is the right term for this kind of operation i want to do by the way?) - any short and hopefully easy description out there?
PS: Mathematica outputs $\displaystyle f(n)=\frac{1}{2}(6 n-(-1)^n+1)+1,n\in\mathbb{N}_0$ as the function i am searching for, but I don't know how to get to this result!
Here is a hint: $$\sum_{i=1}^n ( b + a_i)=(b+a_1)+(b+a_2)+\cdots+(b+a_n)=$$ $$(\underbrace{b+b+\cdots+b}_{n\text{ times}})+(a_1+a_2+\cdots+a_n)=nb+\sum_{i=1}^na_i$$ In your situation, $b=3$, and $a_i=(-1)^{i+1}$ (not $(-1)^i$; do you see why?)
To find a formula for $\displaystyle\sum_{i=1}^n(-1)^{i+1}$, try writing out the first few examples: $$\begin{align} \sum_{i=1}^0(-1)^{i+1}&=0\\ \sum_{i=1}^1(-1)^{i+1}&=(-1)^{1+1}=1\\ \sum_{i=1}^2(-1)^{i+1}&=(-1)^{1+1}+(-1)^{2+1}=1+(-1)=0\\ \sum_{i=1}^3(-1)^{i+1}&=(-1)^{1+1}+(-1)^{2+1}+(-1)^{3+1}=1+(-1)+1=1\\ \end{align}$$ Do you see the pattern? The comments on the question above explain what's going on here.