Math Problem Statement

In the expansion of (x^2 - 6/x)^9, find the term free from x.

Solution

The image shows a mathematical problem where we are asked to find the term free from xx in the expansion of the binomial expression:

(x26x)9\left( x^2 - \frac{6}{x} \right)^9

To solve this, we will expand the binomial expression using the binomial theorem, which is given by:

(a+b)n=k=0n(nk)ankbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

In this case, a=x2a = x^2, b=6xb = -\frac{6}{x}, and n=9n = 9.

The general term in the expansion of (x26x)9(x^2 - \frac{6}{x})^9 is:

Tk=(9k)(x2)9k(6x)kT_k = \binom{9}{k} (x^2)^{9-k} \left( -\frac{6}{x} \right)^k

Simplifying this expression:

Tk=(9k)x2(9k)(1)k6kxkT_k = \binom{9}{k} x^{2(9-k)} (-1)^k \frac{6^k}{x^k}

This further simplifies to:

Tk=(9k)(1)k6kx183kT_k = \binom{9}{k} (-1)^k 6^k x^{18 - 3k}

Now, we are asked to find the term free from xx. For a term to be independent of xx, the exponent of xx must be zero. Therefore, we set the exponent of xx in the general term to zero:

183k=018 - 3k = 0

Solving for kk:

3k=18k=63k = 18 \quad \Rightarrow \quad k = 6

So, the term free from xx occurs when k=6k = 6. Substituting k=6k = 6 into the general term formula:

T6=(96)(1)666T_6 = \binom{9}{6} (-1)^6 6^6

Since (1)6=1(-1)^6 = 1, we have:

T6=(96)66T_6 = \binom{9}{6} 6^6

Now, calculate (96)\binom{9}{6}, which is equal to (93)\binom{9}{3} because of symmetry:

(93)=9×8×73×2×1=84\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

Thus, the term free from xx is:

T6=84×66T_6 = 84 \times 6^6

Finally, calculate 666^6:

66=466566^6 = 46656

Therefore, the term free from xx is:

T6=84×46656=3919104T_6 = 84 \times 46656 = 3919104

So, the constant term is 3,919,1043,919,104.

Would you like further details on this solution?

Here are 5 related questions:

  1. How does the binomial theorem apply to polynomial expansions?
  2. What is the role of the general term in identifying specific terms in an expansion?
  3. Why is the exponent of xx set to zero to find the constant term?
  4. How can you compute binomial coefficients using Pascal’s triangle?
  5. Can the same process be applied to other powers or terms in binomial expansions?

Tip: Binomial expansions can often simplify complex algebraic expressions and are particularly useful when looking for specific terms or behaviors of a function.

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Math Problem Analysis

Mathematical Concepts

Algebra
Binomial Theorem
Exponent Rules

Formulas

Binomial expansion formula: (a + b)^n = Σ binom(n, k) * a^(n-k) * b^k
Exponent rules for simplifying expressions with powers of variables

Theorems

Binomial Theorem

Suitable Grade Level

Grades 10-12