Question

how to prove this?

Answer 1

\[b>a\]
\[b^{n-i}>a^{n-i}\]
\[a^ib^{n-i}>a^ia^{n-i}=a^n\]
\[\sum_{i=0}^{n}a^ib^{n-i}>\sum_{i=0}^{n}a^n\]
\[\frac{b^{n+1}-a^{n+1}}{b-a}>(n+1)a^n\]

Answer 2

how to know \[\sum_{i=0}^{n}a^n = (n+1)a^n\]

Answer 3

\[\sum\limits_{i=0}^{n}a^n=\underbrace{a^n+a^n+\cdots+a^n}_{(n+1)-times}=(n+1)a^n\]

Answer 4

Thanks, you had been a great help =D