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+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

(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

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