Question

# 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.

Solution

## Let P(n) be the given statement, i.e.,  P(n):c(a1+a2+......+an)=ca1+ca2+....+can for all natural numbers n≥2, for c, a1,a2,....anϵR We observe that P(2) is true since c(a1+a2)=ca1+ca2 (by distributive law) Assume that P(n) is true for some natural number k, where k>2, i.e., P(k) c(a1+a2+....+ak)=ca1,ca2+.....cak Now to prove P(k + 1) is true, we have P(k+1);c(a1+a2+...+ak+ak+1) =c((a1+a2+...+ak)+ak+1) =c(a1+a2+....+ak)+cak+1 (By distributive law) =ca1+ca2+....+cak+cak+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≥2.

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