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=\frac{n(n+1)}2`