chứng minh rằng với số nguyên dương n 6 thì số $a_n = 1 + \dfrac{2.6.10....(4n-2)}{(n + 5) (n + 6) ... (2n)}$ là 1 số chính phương

`a_n=1+(2.6.10...(4n-2))/((n+5)(n+6)...(2n))`

`a_n=1+(2^n(1.3.5...(2n-1)))/((n+5)(n+6)...(2n))`

`a_n=1+(2^n[1.2.3.4.5...(2n-1).(2n)])/([2.4...(2n)].(n+5)(n+6)...(2n))`

`a_n=1+(2^n .(2n)!)/([2^n(1.2...n)].(n+5)(n+6)...(2n))`

`a_n=1+(2^n .(2n)!)/(2^n .n!.(n+5)(n+6)...(2n))`

`a_n=1+(n!(n+1)(n+2)...(2n))/(n!(n+5)(n+6)...(2n))`

`a_n=1+(n+1)(n+2)(n+3)(n+4)`

`a_n=1+[(n+1)(n+4)][(n+2)(n+3)]`

`a_n=1+(n^2+4n+n+4)(n^2+3n+2n+6)`

`a_n=1+(n^2+5n+4)(n^2+5n+6)`

`a_n=1+[(n^2+5n+5)-1][(n^2+5n+5)+1]`

`a_n=1+(n^2+5n+5)^2-1`

`a_n=(n^2+5n+5)^2`

`->a_n=1+(2.6.10...(4n-2))/((n+5)(n+6)...(2n))` là `1` số chính phương

`->`đpcm