andrewpriest
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Proof arithmetic mean
- Thread starter andrewpriest
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- Nov 24, 2012
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Hello, and welcome to FMH! 
I would begin with:
[MATH]nS_n=\sum_{i=1}^n\left(x_i\right)[/MATH]
[MATH](n+1)S_{n+1}=\sum_{i=1}^{n+1}\left(x_i\right)[/MATH]
Now, if we subtract the former from the latter, there results:
[MATH](n+1)S_{n+1}-nS_n=x_{n+1}[/MATH]
[MATH](n+1)S_{n+1}=x_{n+1}+nS_n[/MATH]
[MATH](n+1)S_{n+1}=x_{n+1}-S_n+(n+1)S_n[/MATH]
Can you finish?
I would begin with:
[MATH]nS_n=\sum_{i=1}^n\left(x_i\right)[/MATH]
[MATH](n+1)S_{n+1}=\sum_{i=1}^{n+1}\left(x_i\right)[/MATH]
Now, if we subtract the former from the latter, there results:
[MATH](n+1)S_{n+1}-nS_n=x_{n+1}[/MATH]
[MATH](n+1)S_{n+1}=x_{n+1}+nS_n[/MATH]
[MATH](n+1)S_{n+1}=x_{n+1}-S_n+(n+1)S_n[/MATH]
Can you finish?