Question

for every positive integer n, prove that 1+2+...+n= n(n+1)/2

Answer 1

this is just a guess but it has some to do with n!

Answer 2

^thing

Answer 3

so take for instance n=3! so the left will look like this. 3+2+1=3!(3!+1!)/2

Answer 4

i know that i have to use mathematical induction but I can't seen to get past (n+1)(n+2)/2

Answer 5

n!=n(n+1)(n+2)....3+2+1is all i have

Answer 6

thanks but I am really looking for some kind of induction step. That's what we went over in class

Answer 7

Oh wait

Answer 8

I found a webpage with this question here: http://www.purplemath.com/modules/inductn.htm

Answer 9

\[assume 1+2+3+...+k=\frac{ k \left( k+1 \right) }{ 2 } tobe true\]
adding both sides k+1
\[1+2+3+...+k+(k+1)=\frac{ k \left( k+1 \right) }{2 }+\left( k+1 \right)\]
\[=\left( k+1 \right)\left( \frac{ k }{2 }+1 \right)=\frac{ \left( k+1 \right)\left( k+2 \right) }{ 2 }\]
\[=\frac{ \left( k+1 \right)\left( k+1+1 \right) }{2 }\]
which shows if it is true for n=k, it is also true for n=k+1
hence by induction it is true for all n.