%5E-3-expansion.jpeg "(1-x)^-3 Expansion")
Binomial Expansion of (1-x)^-3
In algebra, binomial expansion is a method of expanding an expression of the form $(a+b)^n$, where $a$ and $b$ are variables and $n$ is a positive integer. In this article, we will explore the binomial expansion of $(1-x)^{-3}$.
Binomial Theorem
The binomial theorem is a formula for expanding powers of a binomial expression. It states that
$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$
where $\binom{n}{k}$ is the binomial coefficient, given by
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$
Expansion of (1-x)^-3
Using the binomial theorem, we can expand $(1-x)^{-3}$ as follows:
$(1-x)^{-3} = \sum_{k=0}^{-3} \binom{-3}{k} 1^{-3-k} (-x)^k$
Since the upper limit of the sum is negative, we can rewrite the sum as
$(1-x)^{-3} = \sum_{k=0}^{\infty} \binom{-3}{k} 1^{-3-k} (-x)^k$
Now, we can calculate the binomial coefficients:
$\binom{-3}{0} = 1$ $\binom{-3}{1} = -3$ $\binom{-3}{2} = \frac{(-3)(-4)}{2} = 6$ $\binom{-3}{3} = \frac{(-3)(-4)(-5)}{6} = 10$ $\binom{-3}{k} = \frac{(-3)(-4)(-5)\cdots(-(3+k-1))}{k!}$
Substituting these values back into the sum, we get
$(1-x)^{-3} = 1 -3(-x) + 6(-x)^2 - 10(-x)^3 + \cdots$
Simplifying the expression, we finally get
$(1-x)^{-3} = 1 + 3x + 6x^2 + 10x^3 + \cdots$
This is the binomial expansion of $(1-x)^{-3}$.
Conclusion
In this article, we have seen how to expand $(1-x)^{-3}$ using the binomial theorem. The resulting expansion is an infinite series, where each term is a power of $x$. This expansion is useful in various applications in mathematics and computer science.
