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.
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: $$\sum^{n }_{i=1}ra_i =r\Big(\sum^{n }_{i=1}a_i \Big)$$