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

Question #3886
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:
Expert's answer
Let us expand the expression for unknown n firstly:
(1+x+ax2 )n=(1 + x(1 + ax))n= 1+Cn1 x(1+ax) + Cn2 x2 (1+ax)2 + ... + xn (1+ax)n

Let’s expand second and third terms in the expression we’ve got above:
Cn1 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

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