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Knowing that $(a_i)_{i\ge1}$ prove that $\forall n \in \Bbb N$: $$\sum^n_{i=1}ra_i=r\Big(\sum^n_{i=1}a_i \Big)$$

This kind of demonstrations is totally strange to me, I do not understand how to organize and how to develop it, so I appreciate someone could guide me, recommend books, websites, videos, or any kind of information about it.

Tomi
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  • For how to organize a proof by induction, I like http://math.stackexchange.com/questions/19485/dominoes-and-induction-or-how-does-induction-work/19488#19488 – Ross Millikan Apr 16 '13 at 02:18
  • Something is wrong: there's only $a_i$ in an obvious equality. What about $b_i$? – Przemysław Scherwentke Apr 16 '13 at 02:20
  • My fault; edited. – Tomi Apr 16 '13 at 02:27
  • Are you familiar with the Principle of Induction? I know I had trouble with it when I first learned about it. If you are willing to oblige me, please edit your question by stating the Principle of Induction in your own words. Then we can look at how to use it to guide the proof. – Code-Guru Apr 16 '13 at 02:41

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Base of induction: consider the case $n=1$. The statement takes the form $ra_1=ra_1$, obviously true.

Suppose the statement was proved for $n-1$ terms: that is, we know that $$\sum^{n-1}_{i=1}ra_i=r\Big(\sum^{n-1}_{i=1}a_i \Big)$$ Add $ra_n$ to both sides: $$\sum^{n-1}_{i=1}ra_i+ra_n=r\Big(\sum^{n-1}_{i=1}a_i \Big)+ra_n$$ Use the distributive property of multiplication: $$\sum^{n-1}_{i=1}ra_i+ra_n=r\Big(\sum^{n-1}_{i=1}a_i +a_n \Big)$$ Finally, absorb the additional term into the $\Sigma$ notation for the sum: $$\sum^{n }_{i=1}ra_i =r\Big(\sum^{n }_{i=1}a_i \Big)$$

ˈjuː.zɚ79365
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