Question 144476
The first three terms in the expansion of {{{(1+ay)^n}}}  are {{{1}}}, {{{12y}}}, and {{{68y^2}}}

the fact is that 


{{{(1-ay)^n = 1-nay+ (n(n-1)/2) (ay)^2 + ...}}}

Compare coefficients, 

{{{-na = 12}}}......(1)

{{{(1/2)n(n-1)a^2 = 68}}}......(2)

From (2), we have

{{{n^2a^2 - na^2 = 136}}}......(3)

Plug in {{{-na=12}}} in (3): 

{{{12^2-12^2/n = 136}}}

Solve for {{{n}}}: 

{{{n = 18}}}

Solve for {{{a}}}: 

{{{a = -12/18 = -2/3}}}

Check:

{{{(1+(2/3)y)^18}}}

= {{{1 + C(18,1)(2/3)y + C(18,2)(2/3)^2y^2 + ...}}}