The distributive law from algebra states that for all real numbers, c, $a_{1} a n d a_{2}$ , we have $c (a_{1} + a_{2}) = c a_{1} + c a_{2}$ .

Use this law and mathematical induction to prove that, for all natural numbers, $n \geq 2$ , if c, $a_{1}, a_{2}, . . . . . a_{n}$ are any numbers, then

$c (a_{1} + a_{2} + . . . . + a_{n}) = c a_{1} + c a_{2} + . . . . . . + c a_{n}$ .

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Solution

Let P(n) be the given statement, i.e.,

$P (n) : c (a_{1} + a_{2} + . . . . . . + a_{n}) = c a_{1} + c a_{2} + . . . . + c a_{n}$ for all natural numbers $n \geq 2$ , for c, $a_{1}, a_{2}, . . . . a_{n} ϵ R$

We observe that P(2) is true since

$c (a_{1} + a_{2}) = c a_{1} + c a_{2}$ (by distributive law)

Assume that P(n) is true for some natural number k, where $k > 2$ , i.e.,

$P (k) c (a_{1} + a_{2} + . . . . + a_{k}) = c a_{1}, c a_{2} + . . . . . c a_{k}$

Now to prove P(k + 1) is true, we have $P (k + 1); c (a_{1} + a_{2} + . . . + a_{k} + a_{k + 1})$

$= c ((a_{1} + a_{2} + . . . + a_{k}) + a_{k + 1})$

$= c (a_{1} + a_{2} + . . . . + a_{k}) + c a_{k + 1}$

(By distributive law)

$= c a_{1} + c a_{2} + . . . . + c a_{k} + c a_{k + 1}$

Thus P (k + 1) is true, whenever P(k) is true.

Hence, by the principle of Mathematical Induction, P(n) is true for all natural numbers $n \geq 2$ .

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The distributive law from algebra states that for all real numbers c, a_{1} and a_{2}, we have c (a_{1} + a_{2}) = c a_{1} + c a_{2} . Use this law and mathematical induction to prove that, for all natural numbers, n \geq 2, if c, a_{1}, a_{2}, . . ., a_{n} are any real numbers, then c (a_{1} + a_{2} + . . . + a_{n}) = c a_{1} + c a_{2} + . . . + c a_{n}

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The distributive law from algebra states that for all real numbers, c, a1 and a2, we have c(a1+a2)=ca1+ca2. Use this law and mathematical induction to prove that, for all natural numbers, n≥2, if c, a1,a2,.....an are any numbers, then c(a1+a2+....+an)=ca1+ca2+......+can.