What does $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ mean in induction?

Question

If you're given that $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$

and let's say that prove that $\sum_{k=1}^{n}k+\frac{1}{k+n}=blah$, $n>0$ what does $k$ in itself equal? I mean if you had to put $k$ in $n$ terms.

PS - the exact question is different but I need to understand the idea of it to apply it to the actual question.

Answer 1

You can't get "$k$" in terms of $n$ seeing as $k$ is an indexing variable as already pointed out. I think you want to get a closed form of your partial sum in terms of $n$. In which case, there's really no catch-all trick to do this for any partial sum. For this particular one, you can split the sum since it's a finite sum into $$ \sum_{i=1}^{n} (i + \frac{1}{i+n}) = \sum_{j=1}^{n} j + \sum_{k=1}^{n} \frac{1}{k+n}$$ and you already know what $\sum_{j=1}^{n} j$ equals.

Now we just want to find a closed form for the rightmost sum. Unfortunately, there doesn't seem to be a closed form for this sum.

In general, what I'd try first for these problems is guessing a closed form then proving it by induction after plugging in some values of $n$, or I'd try to work with the terms (e.g. slick telescoping sum tricks). There are other methods out there (e.g. Abel's summation formula).

Answer 2

In such a sum ($\sum_{k=1}^n$k), which simply means $1+2+\dots+n$, $\;k$ is a dummy variable, i.e. a temporary counter which is incremented at each step of the calculation of the sum, and allows us to control at which step we have arrived. As all dummy variables the final result does NOT depend on $k$, and you may very well replace it with another letter without changing the final result.

You can find a similar situation in the notation for a definite integral: for instance, in the equality $\;\int_0^1 x^2\,\mathrm dx=\frac13$, the left-hand side does NOT depend on $x$, which is a dummy variable.

Answer 3

Honestly, I think this is just one of those things where a lightbulb needs to go off in your head. You usually can't explain this kind of symbolism to someone unless they're willing to blindly accept a list of arbitrary rules and then follow the rules. But this kind of explanation only comes easy if the asker has done it before. There's a boot-strap that must happen here. And no one can do it for you the first time.

The best we can do is explain by example. There is a pattern to the list of equations underneath. You have to find it. The $k$ is not important. \begin{align*}\sum_{k=1}^n k &=1+2+3+\cdots+(n-1)+n\\ \sum_{k=1}^n k^2 &=1^2+2^2+3^2+\cdots +(n-1)^2+n^2\\ \sum_{k=1}^n k^3 &=1^3+2^3+3^3+\cdots +(n-1)^3+n^3\\ \sum_{k=1}^n (2k+1) &=(2\cdot 1+1)+(2\cdot 2+1)+(2\cdot 3+1)+\cdots+(2(n-1)+1)+(2n+1)\\ \sum_{k=1}^n (k+1) &=(1+1)+(2+1)+(3+1)+\cdots +((n-1)+1)+(n+1)\\ \sum_{k=1}^n f(k) &=f(1)+f(2)+f(3)+\cdots +f(n-1)+f(n) \end{align*}