I have a question to ask regarding this math question that was given to me. I have no idea on how to start on the question.
The formula $1+2+3+...+n=$$$\ \frac {n(n+1)}2$$ is true for all intergers n>=1
If n is an integer and n>=2, find a formula for the expression 1+2+3+...+(n-1)
The answer for the question is
$$\frac {(n-1)n}2$$
Can i know the steps that I need to take to get the answer?
This is my first time posting here sorry If I have posted wrongly.
You know that $$1+2+...+(n-1) + n = \frac{n(n+1)}{2}$$
Also it is clear that $$1+2+...+(n-1) = \left[1+2+...+(n-1) + n\right] - n$$ Now replacing what is between the $"[\ ]"$ by the formula we know : $$ 1+2+...+(n-1) = \frac{n(n+1)}{2} - n $$ I let you do the algebra to get the result you want.
Note that the "$n$" in the original formula is the last term in the sum.
So the expression for the sum can be thought of as $$\frac{\textrm{(last term)(last term + 1)}}2$$
In the example you are dealing with, the last term is $(n-1)$. So the sum is $$\require{cancel}\frac{(n-1)((n-\cancel1) + \cancel1)}2=\boxed{\frac{(n-1)(n)}2}$$
Note also that since you are told in your example that $n\geq2$, it follows that $n-1\geq1$ (by just subtracting $1$ from both sides), so that it is "legal" to apply the formula since the "last term" is at least $1$.