Prove by induction that $$1+2+3+\cdots+n= \frac{n(n+1)}{2}$$ for all integers greater than or equal to $2$
How can you solve this if the base case is not $1$? I thought it might be a strong induction case but strong induction also requires the initial case. To solve this, should I just assume that the initial value $1$ is assumed?
You can use induction on any infinite sequence of natural numbers (or integers, or any countably infinite sequence at all).
Basic induction says if S is a set of natural numbers and 0 is in S and (s + 1) is in S whenever s is in S then S = N, the whole set of natural numbers.
This is then applied to an inductive proof by letting S be the set of numbers for which P(s) is true, where P is a proposition based on the number s. So if P(0) is true then 0 is in S and if P(s+1) is true whenever P(s) is true then s+1 is in s whenever s is in S and S = N and P(n) is true for all n in N.
Is now T is some countably infinite sequence of $t_0, t_1, ....$ let S be the set of numbers for which $P(t_s)$ is true. If $P(t_0)$ is true and $P(t_{s+1})$ is true whenever $P(t_s)$ is true, then S = N and P(t) is true for all of the sequence T.
In your case the sequence is simply the natural numbers starting from 2 instead of from 0.