Answer to Question #3886 in Combinatorics | Number Theory for Ceejay

Question

If the first three terms of the expansion of (1+x+ax^2)^n are 1+7x+14x^2, Find the values of n and a?
Please take me step by step D:

Answer 1

Let us expand the expression for unknown n firstly:
(1+x+ax² )ⁿ=(1 + x(1 + ax))ⁿ= 1+C_n¹ x(1+ax) + C_n² x² (1+ax)² + ... + xⁿ (1+ax)ⁿ

Let’s expand second and third terms in the expression we’ve got above:
C_n¹ x(1+ax) = nx(1+ax) = nx+anx²
Cn² x² (1+ax)² = n(n-1)/2 x² (1 + 2ax + a² x² ) = n(n-1)/2 x² + an(n-1) x³ + (a² n(n-1))/2 x⁴

Third and all next terms have multiplier x³, thus, they don’t contribute to the first three terms of the expansion of
(1 + x + ax² )ⁿ = 1 + 7x + 14x² + ... given in the problem. Then we may write down following:
(1 + x + ax² )ⁿ=1 + (nx + anx² ) + (n(n - 1)/2 x² + an(n -1) x³ + (a² n(n -1))/2 x⁴ ) + ... = 1 + nx + (an + n(n -1)/2) x²+ ...
Now let’s assign coefficients near the same powers of x:
{ 7x = nx
14x² = (an + n(n -1)/2) x²

From the first equation we see that n=7 and can substitute this into second equation:

14x² = (7a + 7(7 -1)/2) x²
14 = 7a + 21
7a = -7
a = -1
So, we found n = 7 and a = -1.

ANSWER
n = 7, a = -1