The distributive law from algebra states that for real numbers c, a 1 and a 2 , we have c(a 1 + a 2 ) = c a 1 + ca 2 Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 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 .

Question

Philoid · Accepted Answer

Let P(n):c(a₁+a₂+…+a_n) = ca₁+ca₂+…+ca_n ,for all natural

numbers, n ≥ 2.

Step1: For n=2,

P(2)

LHS= c(a₁ + a₂)

RHS= c a₁ + ca₂

As, it is given that c(a₁ + a₂) = c a₁ + ca₂

Thus, P(2) is true.

Step2: For n=k,

Let P(k) be true

So, c(a₁+a₂+…+a_k) = ca₁+ca₂+…+ca_k

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1):

LHS= c(a₁+a₂+…+a_K+ak₊₁)

=c[(a₁+a₂+…+a_K)+a_k+1]

=c(a₁+a₂+…+a_K)+ca_k+1

=ca₁+ca₂+…+ca_K+ca_k+1

=RHS

Thus, P(k+1) is true, so by mathematical induction P(n) is true.

The distributive law from algebra states that for real numbersc, a1 and a2, we have c(a1 + a2) = c a1 + ca2Use this law and mathematical induction to prove that, for allnatural numbers, n ≥ 2, if c, a1, a2, …... an are any real numbers,then c(a1 + a2 +…+ an) = c a1 + c a2 +…+ c an.