Answer:
Step-by-step explanation:
Hello, I will compute twice the sum so I need to compute,
[tex]\begin{aligned}1&+2&+...+&(n-1)&+n\\\\n&+(n-1)&+...+&2&+1\\\\-&-----&---&---&---\\\\(n+1)&+(n+1)&+...+&(n+1)&+(n+1)\end{aligned}[/tex]
1+n = n+1
2 + n-1=n+1
...
n-1+2=n+1
n+1=n+1
So, this is n times n+1 so it is n*(n+1)
[tex]2(1+2+3...+n)=n(n+1)\\\\1+2+3...+n=\dfrac{n(n+1)}{2}[/tex]
