# 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:

Please take me step by step D:

Expert's answer

Let us expand the expression for unknown n firstly:

(1+x+ax

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

C

Cn

Third and all next terms have multiplier x

(1 + x + ax

(1 + x + ax

Now let’s assign coefficients near the same powers of x:

{ 7x = nx

14x

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

14x

14 = 7a + 21

7a = -7

a = -1

So, we found n = 7 and a = -1.

ANSWER

n = 7, a = -1

(1+x+ax

^{2})^{n}=(1 + x(1 + ax))^{n}= 1+C_{n}^{1}x(1+ax) + C_{n}^{2}x^{2}(1+ax)^{2}+ ... + x^{n}(1+ax)^{n}Let’s expand second and third terms in the expression we’ve got above:

C

_{n}^{1}x(1+ax) = nx(1+ax) = nx+anx^{2}Cn

^{2}x^{2}(1+ax)^{2}= n(n-1)/2 x^{2}(1 + 2ax + a^{2}x^{2}) = n(n-1)/2 x^{2}+ an(n-1) x^{3}+ (a^{2}n(n-1))/2 x^{4}Third and all next terms have multiplier x

^{3}, thus, they don’t contribute to the first three terms of the expansion of(1 + x + ax

^{2})^{n}= 1 + 7x + 14x^{2}+ ... given in the problem. Then we may write down following:(1 + x + ax

^{2})^{n}=1 + (nx + anx^{2}) + (n(n - 1)/2 x^{2}+ an(n -1) x^{3}+ (a^{2}n(n -1))/2 x^{4}) + ... = 1 + nx + (an + n(n -1)/2) x^{2}+ ...Now let’s assign coefficients near the same powers of x:

{ 7x = nx

14x

^{2 }= (an + n(n -1)/2) x^{2}From the first equation we see that n=7 and can substitute this into second equation:

14x

^{2}= (7a + 7(7 -1)/2) x^{2}14 = 7a + 21

7a = -7

a = -1

So, we found n = 7 and a = -1.

ANSWER

n = 7, a = -1

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