The distributive law from algebra states that for real numbers c, a1

1.	The distributive law from algebra states that for 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 real numbers, then c(a1 + a2 +…+ an) = ca1 + ca2 +…+ can.
Answer» Let P(n):c(a₁+a₂+…+an) = 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+a_k+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 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 real numbers, then c(a1 + a2 +…+ an) = ca1 + ca2 +…+ can.

Answer»

Let P(n):c(a₁+a₂+…+an) = 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+a_k+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.