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Fundamental Theorem of Arithmetic

n1111+n55+n33...

Question

$\frac{n^{11}}{11} + \frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{62}{165} n$ is a positive integer for all n ∈ N.

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Solution

Let P(n) be the given statement.
Now,
$P (n) : \frac{n^{11}}{11} + \frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{62}{165} n is a positive integer for all n \in N . Step 1 : P (1) = \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{62}{165} = \frac{15 + 33 + 55 + 62}{165} = \frac{165}{165} = 1 It is certainly a positive integer . Hence, P (1) is true . Step 2 : Le t P (m) be true . Then, \frac{m^{11}}{11} + \frac{m^{5}}{5} + \frac{m^{3}}{3} + \frac{62}{165} m is a positive integer . Now, let \frac{m^{11}}{11} + \frac{m^{5}}{5} + \frac{m^{3}}{3} + \frac{62}{165} m = λ, where λ \in N is a positive integer . We have to show that P (m + 1) is true whenever P (m) is true . To prove : \frac{(m + 1)^{11}}{11} + \frac{(m + 1)^{5}}{5} + \frac{(m + 1)^{3}}{3} + \frac{62}{165} (m + 1) is a positive integer . Now, \frac{(m + 1)^{11}}{11} + \frac{(m + 1)^{5}}{5} + \frac{(m + 1)^{3}}{3} + \frac{62}{165} (m + 1) = \frac{1}{11} (m^{11} + 11 m^{10} + 55 m^{9} + 165 m^{8} + 330 m^{7} + 462 m^{6} + 462 m^{5} + 330 m^{4} + 165 m^{3} + 55 m^{2} + 11 m + 1) + \frac{1}{5} (m^{5} + 5 m^{4} + 10 m^{3} + 10 m^{2} + 5 m + 1) + \frac{1}{3} (m^{3} + 3 m^{2} + 3 m + 1) + \frac{62}{165} m + \frac{62}{165} = [\frac{m^{11}}{11} + \frac{m^{5}}{5} + \frac{m^{3}}{3} + \frac{62}{165} m] + m^{10} + 5 m^{9} + 15 m^{8} + 30 m^{7} + 42 m^{6} + 42 m^{5} + 31 m^{4} + 17 m^{3} + 8 m^{2} + 3 m + \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{6}{105} = λ + m^{10} + 5 m^{9} + 15 m^{8} + 30 m^{7} + 42 m^{6} + 42 m^{5} + 31 m^{4} + 17 m^{3} + 8 m^{2} + 3 m + 1 It is a positive integer . Thus, P (m + 1) is true . By the principle of mathematical induction, P (n) is true for all n \in N .$

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