The distributive law from algebra states that for all real numbers

Solution:

Given: For all real numbers $c, a_{1}$ and $a_{2}, c\left(a_{1}+a_{2}\right)=c a_{1}+c a_{2}$.

To prove: For all natural numbers, $n \geq 2$, if $c, a_{1}, a_{2}, \ldots, a_{n}$ are any real numbers, then $c\left(a_{1}+a_{2}+\ldots+a_{n}\right)=c a_{1}+c a_{2}+\ldots+c a_{n}$

Proof:

Let $\mathrm{P}(n): c\left(a_{1}+a_{2}+\ldots+a_{n}\right)=c a_{1}+c a_{2}+\ldots+c a_{n}$ for all natural numbers $n \geq 2$ and $c, a_{1}$, $a_{2}, \ldots, a_{n} \in \mathbf{R}$.

Step I: For $n=2$,

$\mathrm{P}(2)$ :

LHS $=c\left(a_{1}+a_{2}\right)$

RHS $=c a_{1}+c a_{2}$

As, $c\left(a_{1}+a_{2}\right)=c a_{1}+c a_{2} \quad$ (Given)

$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

So, it is true for $n=2$.

Step II : For $n=k$,

Let $\mathrm{P}(k): c\left(a_{1}+a_{2}+\ldots+a_{k}\right)=c a_{1}+c a_{2}+\ldots+c a_{k}$ be true for some natural numbers $k \geq 2$ and $c, a_{1}, a_{2}, \ldots, a_{k} \in \mathbf{R}$.

Step III : For $n=k+1$,

$\mathrm{P}(k+1):$

$\mathrm{LHS}=c\left(a_{1}+a_{2}+\ldots+a_{k}+a_{k+1}\right)$

$=c\left[\left(a_{1}+a_{2}+\ldots+a_{k}\right)+a_{k+1}\right]$

$=c\left(a_{1}+a_{2}+\ldots+a_{k}\right)+c a_{k+1}$

$=c a_{1}+c a_{2}+\ldots+c a_{k}+c a_{k+1}$ (Using step II)

$\mathrm{RHS}=c a_{1}+c a_{2}+\ldots+c a_{k}+c a_{k+1}$

As, LHS $=$ RHS

So, it is also true for $n=k+1$.

Hence, for all natural numbers, $n \geq 2$, if $c, a_{1}, a_{2}, \ldots, a_{n}$ are any real numbers, then $c\left(a_{1}+a_{2}+\ldots+a_{n}\right)=c a_{1}+c a_{2}+\ldots+c a_{n}$.