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Question
Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.
Solution
Let P(n): `n^5/5 + n^3/3 + (7n)/15`, ∀ n ∈ N.
Step 1: P(1): `1^5/5 + 1^3/3 + (71)/15`
= `(3 + 5 + 7)/15`
= `15/13`
= 1 ∈ N
Which is true for P(1).
Step 2: P(k): `k^5/5 + k^3/3 + (7.k)/15`
Let it be true for P(k) and let `k^5/5 + k^3/3 + (7k)/15` = λ.
Step 3: P(k + 1) = `(k + 1)^5/5 + (k + 1)^3/3 + (7(k + 1))/15`
= `1/5 [k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1] + 1/3 [k^3 + 3k^2 + 3k + 1]`
= `(k^5/5 + k^3/3 + (7k)/15) + (k^4 + 2k^3 + 2k) + 1/5 + 1/3 + 7/15 + 71/15 k + 7/15`
= `lambda + k^4 + 2k^3 + 3k^2 + 2k + 1` ....[From step 2]
= Positive integers
= Natural number
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.
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