Question

Show that sum k=1/2n(n+1) from k=1 to n

Answer 1

Suppose the value of the sum is $S$:

$S=\displaystyle\sum_{k=1}^nk$

So

$S=1+2+3+\cdots+(n-2)+(n-1)+n$

but also

$S=n+(n-1)+(n-2)+\cdots+3+2+1$

That is,

$S=\displaystyle\sum_{k=1}^n(n-k+1)$

Adding these together, we have

$2S=\displaystyle\sum_{k=1}^n(k+n-k+1)=\sum_{k=1}^n(n+1)=(n+1)\sum_{k=1}^n1=n(n+1)$
$\implies S=\dfrac{n(n+1)}2$