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The discussion centers on proving that each term of the sequence defined by a_n = 160{n+6 choose 6} is an integer for n ≥ 3. The proof uses mathematical induction, starting with the base case where a_3 is calculated and shown to be an integer. The inductive step demonstrates that if the formula holds for n, it also holds for n+1, reinforcing the integer property. Corrections to the original formula were acknowledged, confirming the validity of the approach. The conclusion emphasizes that the result follows from the properties of binomial coefficients, which are inherently integers.
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$a_1=3,a_2=5757,\,\, a_n=\dfrac {7(a_a+a_2+-------+a_n)}{n}, \,\, (n\geq2)\,\, prove \,\,each\,\, term\,\, of\,\, a_n\,\,
is \,\, an \,\, integer$

correction :
$a_1=3,a_2=5757,\,\, a_n=\dfrac {7(a_1+a_2+-------+a_{n-1})}{n}, \,\, (n\geq2)\,\, prove \,\,each\,\, term\,\, of\,\, a_n\,\,
is \,\, an \,\, integer$
 
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Albert said:
$a_1=3,a_2=5757,\,\, a_n=\dfrac {7(a_{\color{red}1}+a_2+-------+a_{n{\color{red}-1}})}{n}, \,\, (n\geq2)\,\, prove \,\,each\,\, term\,\, of\,\, a_n\,\,
is \,\, an \,\, integer$
[sp]Assuming that I'm right in making those corrections to the question, I claim that $$a_n = 160{n+6\choose6}$$ 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]
 
Opalg said:
[sp]Assuming that I'm right in making those corrections to the question, I claim that $$a_n = 160{n+6\choose6}$$ 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 !
 
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