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Question
The distributive law from algebra says 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 real numbers, then c(a1 + a2 + ... + an) = ca1 + ca2 + ... + can.
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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, ... an ∈ R.
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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