Math Problem Statement

In the expansion of (32x)(1+x2)n\left(3-2x\right)\left(1+\frac{x}{2}\right)^{n}. The coefficient of x is 7. Find the value of the constant n and hence find the coefficient of x2x^2

Solution

We are tasked with expanding (32x)(1+x2)n\left(3 - 2x\right)\left(1 + \frac{x}{2}\right)^{n} and finding the value of nn such that the coefficient of xx is 7. Afterward, we will also determine the coefficient of x2x^2.

Step 1: Expand (1+x2)n\left(1 + \frac{x}{2}\right)^{n}

We will use the binomial theorem to expand (1+x2)n\left(1 + \frac{x}{2}\right)^{n} up to the terms involving xx and x2x^2, since we're interested in the coefficients of xx and x2x^2. The binomial expansion is:

(1+x2)n=1+(n1)(x2)+(n2)(x2)2+\left(1 + \frac{x}{2}\right)^{n} = 1 + \binom{n}{1} \left(\frac{x}{2}\right) + \binom{n}{2} \left(\frac{x}{2}\right)^2 + \cdots

This simplifies to:

(1+x2)n=1+nx2+n(n1)2(x2)2+\left(1 + \frac{x}{2}\right)^{n} = 1 + n\frac{x}{2} + \frac{n(n-1)}{2} \left(\frac{x}{2}\right)^2 + \cdots

(1+x2)n=1+n2x+n(n1)8x2+\left(1 + \frac{x}{2}\right)^{n} = 1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots

Step 2: Multiply by 32x3 - 2x

Now, we multiply the expansion of (1+x2)n\left(1 + \frac{x}{2}\right)^{n} by 32x3 - 2x:

(32x)(1+n2x+n(n1)8x2+)(3 - 2x)\left(1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots\right)

Distribute 32x3 - 2x:

=3(1+n2x+n(n1)8x2+)2x(1+n2x+n(n1)8x2+)= 3(1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots) - 2x(1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots)

This simplifies to:

=3+3n2x+3n(n1)8x22xnx2+= 3 + \frac{3n}{2}x + \frac{3n(n-1)}{8}x^2 - 2x - n x^2 + \cdots

Step 3: Collect terms involving xx

From the expansion, the terms involving xx are:

3n2x2x\frac{3n}{2}x - 2x

For the coefficient of xx to be 7, we set up the equation:

3n22=7\frac{3n}{2} - 2 = 7

Solve for nn:

3n2=9\frac{3n}{2} = 9

3n=183n = 18

n=6n = 6

Step 4: Find the coefficient of x2x^2

Now that we know n=6n = 6, substitute this into the terms involving x2x^2. The terms involving x2x^2 are:

3n(n1)8x2nx2\frac{3n(n-1)}{8}x^2 - n x^2

Substitute n=6n = 6:

3×6×58x26x2\frac{3 \times 6 \times 5}{8}x^2 - 6x^2

Simplify:

908x26x2=11.25x26x2=5.25x2\frac{90}{8}x^2 - 6x^2 = 11.25x^2 - 6x^2 = 5.25x^2

Thus, the coefficient of x2x^2 is 5.255.25.

Final Answer:

  • The value of nn is 66.
  • The coefficient of x2x^2 is 5.255.25.

Would you like further details on any step?

Here are five related questions for practice:

  1. What is the binomial expansion of (1+x3)5\left(1 + \frac{x}{3}\right)^5?
  2. How would the expansion change if the expression was (2x)(1+x3)n(2 - x)\left(1 + \frac{x}{3}\right)^n?
  3. What is the general form of the binomial expansion for (a+bx)n(a + bx)^n?
  4. How do you find the coefficient of a specific power in a binomial expansion?
  5. What are the properties of binomial coefficients?

Tip: In binomial expansions, identifying the terms you need early on can save a lot of time.

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Math Problem Analysis

Mathematical Concepts

Algebra
Binomial Expansion
Polynomial Coefficients

Formulas

Binomial expansion: (1 + x)^n = 1 + nx + n(n-1)/2 * x^2 + ...
Polynomial multiplication: Distribute terms across (3 - 2x) and the expanded binomial.

Theorems

Binomial Theorem

Suitable Grade Level

Grades 10-12