Question:
how the procedure in mathematical induction is true to prove that any number is divisible by any other number.for eg. we have to prove
1 3 5.............n is divisble by 10 then we get in every similar question like 10a where a= Z(Integers) now question is we can fill any integer and somehow get 10a.
Answer:
Let us take an example:
Ex: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) is divisible by (1/3){n(n + 1)(n + 2)}
Solution:
Let the given statement be P(n). Then,
P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k + 1) = (1/3){k(k + 1)(k + 2)}.
Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +...+ k(k + 1) + (k + 1)(k + 2)
= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ....... + k(k + 1)) + (k + 1)(k + 2)
= (1/3) k(k + 1)(k + 2) + (k + 1)(k + 2) [using (i)]
= (1/3) [k(k + 1)(k + 2) + 3(k + 1)(k + 2)
= (1/3){(k + 1)(k + 2)(k + 3)}
⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +......+ (k + 1)(k + 2)
= (1/3){k + 1 )(k + 2)(k +3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values of ∈ N.
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