Prove that n^ 11 11 + n^5 5 + n^3 3 +62 165 n is a positive integer for all n .

Let p(n): n^ 11 11 + n^5 5 + n^3 3 +62 165 n is a positive integer For n = 1 1 11 +1 5 +1 3 +62 165 =15+33+55+62 165 =165 165 Which is a positive integer Let p(n) is true for n = k, So k^ 11 11 + k^5 5 + k^3 3 +62 165 is positive integer k^ 11 11 + k^5 5 + k^3 3 +62 165 = ...(1) For n = k + 1, (k+1)^ 11 11 + (k+1)^5 5 + (k+1)^3 3 +62 165 (k+1) =1 11 [k^ 11 +11k^ 10 +55k^9+165k^8+330k^7+462k^6 +462k^5+330k^4+165k^3+55k^2+11k+1 ] +1 5 [k^5+5k^4+10k^3+10k^2+5k+1 ] +1 3 [k^3+3k^2+3k+1 ]+62 165 [k+1 ] = [ k^ 11 11 + k^5 5 + k^3 3 +62 165 ]+ [k^ 10 +5k^9+15k^8+30k^7+42k^6+42k^5+ 30k^4+15k^3+5k^2+1+1 11 +k^4+2k^3+2k^2+1 5 +k^2+k+1 3 +62 165 ] = +k^ 10 +5k^9+15k^8+30k^7+42k^6+42k^5+31k^4+17k^3+8k^3+2k +1 = +k^6+3k^5+6k^4+7k^3+6k^2+3k+1 = An integer ⇒ p(n) is true for n = k + 1 ⇒ p(n) is true for all by n PMI

Prove that n^ 11 11 + n^5 5 + n^3 3 +62 165 n is a positive integ…

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Question

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

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Answer

Let p(n): $\frac{\text{n}^{11}}{11}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{62}{165}$ n is a positive integer
For n = 1
$\frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{62}{165}$
$=\frac{15+33+55+62}{165}$
$=\frac{165}{165}$
Which is a positive integer
Let p(n) is true for n = k, So
$\frac{\text{k}^{11}}{11}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{62}{165}$ is positive integer
$\frac{\text{k}^{11}}{11}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{62}{165}=\lambda \ ...(1)$
For n = k + 1,
$\frac{(\text{k+1})^{11}}{11}+\frac{(\text{k+1})^5}{5}+\frac{(\text{k+1})^3}{3}+\frac{62}{165}(\text{k+1})$
$=\frac{1}{11}\big[\text{k}^{11}+11\text{k}^{10}+55\text{k}^9+165\text{k}^8+330\text{k}^7+462\text{k}^6\\+462\text{k}^5+330\text{k}^4+165\text{k}^3+55\text{k}^2+11\text{k}+1\big]\\+\frac{1}{5}\big[\text{k}^5+5\text{k}^4+10\text{k}^3+10\text{k}^2+5\text{k}+1\big]\\+\frac{1}{3}\big[\text{k}^3+3\text{k}^2+3\text{k}+1\big]+\frac{62}{165}\big[\text{k}+1\big]$
$=\Big[\frac{\text{k}^{11}}{11}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{62}{165}\Big]+\Big[\text{k}^{10}+5\text{k}^9+15\text{k}^8+30\text{k}^7+42\text{k}^6+42\text{k}^5+\\30\text{k}^4+15\text{k}^3+5\text{k}^2+1+\frac{1}{11}+\text{k}^4+2\text{k}^3+2\text{k}^2+\frac{1}{5}+\text{k}^2+\text{k}+\frac{1}{3}+\frac{62}{165}\Big]$
$=\lambda+\text{k}^{10}+5\text{k}^9+15\text{k}^8+30\text{k}^7+42\text{k}^6+42\text{k}^5+31\text{k}^4+17\text{k}^3+8\text{k}^3+2\text{k} +1$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+1$
= An integer
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI

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