$\sum_{k=1}^m n-k$ and $\sum_{k=m+1}^{2m} k$

Question

i really can't find out why

$$\sum_{k=1}^m n-k = -\frac12\left(m^2-2nm +m\right)$$ and why $$\sum_{k=m+1}^{2m} k = \frac12m\, (3m +1)$$

For the first one i really don't know where to start, but for the second one i tried to put the numbers one near the other in this way and under the opposite sequence \begin{align*}m+1, m+2,&\dots,\, 2m-1, 2m\\ 2m,2m-1,&\dots,\,m+2,m+1\end{align*}

I saw that the vertical sum is always $3m+1$, that it must be multiplied for the number of numbers in a sequence, that is $m$, and that divided by $2$, so $$\dfrac{(3m+1)m}{2}$$ But i don't know if it can be considered as a proof. I tried to prove similary the first one with this method, but it doesn't work, do you have any hint?

Answer 1

For the first one:

$$ \sum_{k=1}^m (n-k)=(n-1)+(n-2)+(n-3)+ \dots + (n-(m-1))+(n-m)=mn-(1+2+ \dots +m)=mn- \frac{m(m+1)}{2}=mn-\frac{m^2+m}{2}=\frac{2mn-m^2-m}{2}=-\frac{1}{2}(m^2-2mn+m)$$

For the second one:

Set $u=k-m$ When $k=m+1 \Rightarrow u=1$

When $k=2m \Rightarrow u=m$

So,

$$\sum_{k=m+1}^{2m} k= \sum_{u=1}^{m} (u+m)$$

Answer 2

$$\sum_{k=1}^m n-k = mn-\sum_{k=1}^m k = mn-\frac{1}{2}m(m-1)= -\frac12\left(m^2-2nm +m\right)$$

I will leave it as an exercise for you to figure out the 2nd one. You many want to write out the terms to see what pattern there is