Thank you! I'm glad it was helpful.

[Albert1]

$a_1=3,a_2=5757,\,\, a_n=\dfrac {7(a_a+a_2+-------+a_n)}{n}, \,\, (n\geq2)\,\, prove \,\,each\,\, term\,\, of\,\, a_n\,\,
is \,\, an \,\, integer$

correction :
$a_1=3,a_2=5757,\,\, a_n=\dfrac {7(a_1+a_2+-------+a_{n-1})}{n}, \,\, (n\geq2)\,\, prove \,\,each\,\, term\,\, of\,\, a_n\,\,
is \,\, an \,\, integer$

 

[Opalg]

$a_1=3,a_2=5757,\,\, a_n=\dfrac {7(a_{\color{red}1}+a_2+-------+a_{n{\color{red}-1}})}{n}, \,\, (n\geq2)\,\, prove \,\,each\,\, term\,\, of\,\, a_n\,\,
is \,\, an \,\, integer$

[sp]Assuming that I'm right in making those corrections to the question, I claim that \(\displaystyle a_n = 160{n+6\choose6}\) for all $n\geqslant3$. The result then follows because binomial coefficients are integers.

To prove the claimed result by induction, check first that $a_1+a_2 = 3+5757 = 5760 = 2^7\cdot3^2\cdot5$, so that $a_3 = \dfrac{2^7\cdot3^2\cdot5\cdot7}3 = 2^7\cdot3\cdot5\cdot7.$ But $$160{9\choose6} = 2^5\cdot5\cdot\frac{9\cdot8\cdot7}{1\cdot2\cdot3} = 2^7\cdot3\cdot5\cdot7.$$ That establishes the base case $n=3$.

Now suppose that the result is true for $n$. Since $a_n = \frac7n(a_1+a_2 + \ldots + a_{n-1})$, it follows that $$a_1+a_2 + \ldots + a_{n-1} = \frac{na_n}7 = \frac{160n}7{n+6\choose6},$$ and so $$a_1+a_2 + \ldots + a_n = \frac{160n}7{n+6\choose6} + 160{n+6\choose6} = \frac{160(n+7)}7{n+6\choose6}.$$ Therefore $$a_{n+1} = \frac7{n+1}(a_1+a_2 + \ldots + a_n) = \frac{160(n+7)}{n+1}{n+6\choose6} = \frac{160(n+7)\cdot (n+6)!}{(n+1)\cdot6!\cdot n!} = \frac{160(n+7)!}{6!(n+1)!} = 160{n+7\choose n+1} = 160{n+7\choose 6}.$$ That completes the inductive step.[/sp]

 

[Albert1]

[sp]Assuming that I'm right in making those corrections to the question, I claim that \(\displaystyle a_n = 160{n+6\choose6}\) for all $n\geqslant3$. The result then follows because binomial coefficients are integers.

To prove the claimed result by induction, check first that $a_1+a_2 = 3+5757 = 5760 = 2^7\cdot3^2\cdot5$, so that $a_3 = \dfrac{2^7\cdot3^2\cdot5\cdot7}3 = 2^7\cdot3\cdot5\cdot7.$ But $$160{9\choose6} = 2^5\cdot5\cdot\frac{9\cdot8\cdot7}{1\cdot2\cdot3} = 2^7\cdot3\cdot5\cdot7.$$ That establishes the base case $n=3$.

Now suppose that the result is true for $n$. Since $a_n = \frac7n(a_1+a_2 + \ldots + a_{n-1})$, it follows that $$a_1+a_2 + \ldots + a_{n-1} = \frac{na_n}7 = \frac{160n}7{n+6\choose6},$$ and so $$a_1+a_2 + \ldots + a_n = \frac{160n}7{n+6\choose6} + 160{n+6\choose6} = \frac{160(n+7)}7{n+6\choose6}.$$ Therefore $$a_{n+1} = \frac7{n+1}(a_1+a_2 + \ldots + a_n) = \frac{160(n+7)}{n+1}{n+6\choose6} = \frac{160(n+7)\cdot (n+6)!}{(n+1)\cdot6!\cdot n!} = \frac{160(n+7)!}{6!(n+1)!} = 160{n+7\choose n+1} = 160{n+7\choose 6}.$$ That completes the inductive step.[/sp]

sorry a typo
Yes you are right in making those corrections to the question
very good solution !