Simple Sum

Replying to the challenge master note. This proof can be completed through mathematical induction.

Prove: $1+2+3+4+...+n=\frac { n(n+1) }{ 2 }$

Show for n=1

$\frac { 1(1+1) }{ 2 } =\frac { 2 }{ 2 } = 1$

Inductive hypothesis. Assume true for n=k

$1+2+3+4+...+k=\frac { k(k+1) }{ 2 }$

Prove for n=k+1

$1+2+3+4+...+k+(k+1)=\frac { (k+1)(k+2) }{ 2 }$

$\frac { k(k+1) }{ 2 } + (k+1) =\frac { (k+1)(k+2) }{ 2 }$

$\frac { { k }^{ 2 }+k }{ 2 } +\frac { 2k+2 }{ 2 } =\frac { (k+1)(k+2) }{ 2 }$

$\frac { { k }^{ 2 }+3k+2 }{ 2 } =\frac { { k }^{ 2 }+3k+2 }{ 2 }$

Q.E.D

Use \displaystyle before the sum to make it look like this: $\displaystyle\sum _{ n=0 }^{ k }{ n } =\frac { k(k+1) }{ 2 }$