First slide

Introduction to P.M.I

Question

For $n \in N, (\frac{1}{5}) n^{5} + (\frac{1}{3}) n^{3} + (\frac{7}{15}) n$ is

Easy

A

an integer
B

a natural number
C

a positive fraction
D

None of the above

Solution

$\begin{array}{l} Let P (n) : (\frac{1}{5}) n^{5} + (\frac{1}{3}) n^{3} + (\frac{7}{15}) n \\ ∴ P (1) = (\frac{1}{5}) (1)^{5} + (\frac{1}{3}) (1)^{3} + (\frac{7}{15}) 1 \\ = \frac{1}{5} + \frac{1}{3} + \frac{7}{15} = \frac{3 + 5 + 7}{15} = \frac{15}{15} = 1 \in N \\ P (2) = (\frac{1}{5}) (2)^{5} + (\frac{1}{3}) (2)^{3} + (\frac{7}{15}) 2 \\ = \frac{32}{5} + \frac{8}{3} + \frac{14}{15} \\ = \frac{96 + 40 + 14}{15} = \frac{150}{15} = 10 \in N \end{array}$
By mathematical induction P(n) is a natural number

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