Question 172778
First three terms are 1, 12y and 68y^2
Your hint says (1+ay)^n=1+nay+n(n-1)/2(ay)^2+
so (1+ay)^n=1+12y+68y^2+...
12y=nay, so na=12 and a=12/n
68y^2 = (n(n-1)/2)(ay)^2 = (n(n-1)/2)a^2y^2. so (n(n-1)/2)a^2 = 68

Now substitute the value for a into the second equation and solve
{{{n(n-1)a^2/2 = 68}}}
{{{((n^2-n)/2) * (12/n)^2 = 68}}}
{{{(n^2-n)(144/2) / (n^2) = 68 }}}
{{{(n^2 -n)72 = 68n^2}}}
{{{72n^2 - 72n = 68n^2}}}
{{{4n^2 = 72n}}}
{{{4n = 72}}}
{{{n = 18}}}
so a = 2/3

now check against the hint to see if the values work
a*n = (2/3)*18 = 12. so far so good
(18*17/2)*(4/9) = 68