Math Problem Statement

Find the first five terms, in ascending powers of 𝑥, in the expansion of (1 + 𝑥)^7

Solution

To find the first five terms in the expansion of (1+x)7(1 + x)^7, we can use the Binomial Theorem, which states that:

(1+x)n=k=0n(nk)xk(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k

Where:

  • (nk)\binom{n}{k} is the binomial coefficient, given by (nk)=n!k!(nk)!\binom{n}{k} = \frac{n!}{k!(n-k)!}.
  • n=7n = 7 in this case.

We'll calculate the first five terms (i.e., up to x4x^4).

  1. For k=0k = 0: (70)x0=1\binom{7}{0} x^0 = 1

  2. For k=1k = 1: (71)x1=7x\binom{7}{1} x^1 = 7x

  3. For k=2k = 2: (72)x2=7×62x2=21x2\binom{7}{2} x^2 = \frac{7 \times 6}{2} x^2 = 21x^2

  4. For k=3k = 3: (73)x3=7×6×53×2×1x3=35x3\binom{7}{3} x^3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} x^3 = 35x^3

  5. For k=4k = 4: (74)x4=7×6×5×44×3×2×1x4=35x4\binom{7}{4} x^4 = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} x^4 = 35x^4

Thus, the first five terms in the expansion of (1+x)7(1 + x)^7 are:

1+7x+21x2+35x3+35x41 + 7x + 21x^2 + 35x^3 + 35x^4

Would you like more details or have any questions? Here are five related questions to explore further:

  1. How is the binomial coefficient calculated in general?
  2. What is the full expansion of (1+x)7(1 + x)^7?
  3. What would the expansion look like for negative exponents, such as (1+x)7(1 + x)^{-7}?
  4. How can the binomial theorem be applied to higher-order powers, like (1+x)10(1 + x)^{10}?
  5. What is the general formula for the expansion of (a+b)n(a + b)^n?

Tip: The binomial theorem is a powerful tool for expanding expressions of the form (1+x)n(1 + x)^n, and it is used extensively in algebra and calculus for approximation techniques!

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Math Problem Analysis

Mathematical Concepts

Algebra
Binomial Expansion
Polynomial Expansions

Formulas

Binomial Theorem: (1 + x)^n = Σ (n choose k) x^k

Theorems

Binomial Theorem

Suitable Grade Level

Grades 10-12