Math Problem Statement

In the expansion of X to the power of seven parentheses X to the power of 4/2+ K over X to the power three parentheses to the power of seven the constant term is 168 find the value of K

Solution

We are asked to find the value of KK in the expansion of:

(x4/2+Kx3)7\left( x^4/2 + \frac{K}{x^3} \right)^7

where the constant term in the expansion is 168.

Step 1: Apply the Binomial Theorem

Using the Binomial Theorem, the expansion of (a+b)n(a + b)^n is given by:

(a+b)n=r=0n(nr)anrbr(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r

In our case, a=x4/2a = x^4/2 and b=Kx3b = \frac{K}{x^3}, and n=7n = 7.

Thus, the general term in the expansion is:

Tr=(7r)(x42)7r(Kx3)rT_r = \binom{7}{r} \left( \frac{x^4}{2} \right)^{7-r} \left( \frac{K}{x^3} \right)^r

Simplifying this:

Tr=(7r)(x4(7r)27r)(Krx3r)T_r = \binom{7}{r} \left( \frac{x^{4(7-r)}}{2^{7-r}} \right) \left( \frac{K^r}{x^{3r}} \right)

Tr=(7r)Kr27rx4(7r)3rT_r = \binom{7}{r} \frac{K^r}{2^{7-r}} \cdot x^{4(7-r) - 3r}

Tr=(7r)Kr27rx287rT_r = \binom{7}{r} \frac{K^r}{2^{7-r}} \cdot x^{28 - 7r}

Step 2: Find the constant term

For the term to be constant, the exponent of xx must be zero. So, we set:

287r=028 - 7r = 0

Solving for rr:

r=4r = 4

Step 3: Substitute r=4r = 4 into the general term

Now, substitute r=4r = 4 into the general expression for TrT_r:

T4=(74)K4274x287(4)T_4 = \binom{7}{4} \frac{K^4}{2^{7-4}} \cdot x^{28 - 7(4)}

T4=(74)K423x0T_4 = \binom{7}{4} \frac{K^4}{2^3} \cdot x^0

Since x0=1x^0 = 1, the constant term is:

T4=(74)K48T_4 = \binom{7}{4} \frac{K^4}{8}

We know that the constant term is given as 168, so:

(74)K48=168\binom{7}{4} \frac{K^4}{8} = 168

Step 4: Solve for KK

First, calculate (74)\binom{7}{4}:

(74)=7!4!(74)!=7×6×53×2×1=35\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

Substitute this into the equation:

35K48=16835 \cdot \frac{K^4}{8} = 168

Multiply both sides by 8:

35K4=168×8=134435 \cdot K^4 = 168 \times 8 = 1344

Now, divide by 35:

K4=134435=38.4K^4 = \frac{1344}{35} = 38.4

Finally, take the fourth root of both sides:

K=38.44K = \sqrt[4]{38.4}

Step 5: Calculate the fourth root

Now, let's compute K=38.44K = \sqrt[4]{38.4}.It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can easily calculate 38.44\sqrt[4]{38.4} using a calculator or an online tool to get the value of KK.

Let me know if you'd like further clarification!

Here are five related questions:

  1. How does the binomial theorem apply to polynomial expansions?
  2. Can you explain the importance of the binomial coefficient in binomial expansions?
  3. What are the properties of exponents in the context of binomial expansions?
  4. How can you determine which term in a binomial expansion contains the constant term?
  5. How do fourth roots behave in more complex equations?

Tip: When solving problems with powers or roots, always ensure your calculator is set to the correct mode for precision.

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Math Problem Analysis

Mathematical Concepts

Binomial Expansion
Exponentiation
Constant Term Identification

Formulas

(a + b)^n = ∑ (n choose r) a^(n-r) b^r
General term: Tr = (n choose r) a^(n-r) b^r
Exponent simplification: x^(4(7-r)) / x^(3r) = x^(28 - 7r)

Theorems

Binomial Theorem

Suitable Grade Level

Grades 10-12