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Chapter 14: Problem 25

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$(x-1)^{5}$$

Short Answer

Expert verified
\((x-1)^{5} = x^{5} - 5x^{4} + 10x^{3} - 10x^{2} + 5x - 1\)

Step by step solution

01

Identification of terms

The first step is to identify the terms and the power in the binomial. Here, \(a=x\), \(b=-1\) and the power \(n=5\).
02

Expansion Using Binomial Theorem

The second step is to apply the binomial theorem, which is \((a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k}b^k\). Applying Binomial Theorem in this case becomes: \((x-1)^{5} = \sum_{k=0}^{5} {5 \choose k} x^{5-k}(-1)^k\)
03

Simplify Each Term

Now, calculate each term in the sum: \((x-1)^{5} = {5 \choose 0} x^{5}(-1)^0 + {5 \choose 1} x^{4}(-1)^1 + {5 \choose 2} x^{3}(-1)^2 + {5 \choose 3} x^{2}(-1)^3 + {5 \choose 4} x^{1}(-1)^4 + {5 \choose 5} x^{0}(-1)^5\)
04

Compute Each Term

Now, compute each term:\n\({5 \choose 0} x^{5}(-1)^0 = 1*x^{5} = x^{5}\), \n\({5 \choose 1} x^{4}(-1)^1 = 5*x^{4}*(-1) = -5x^{4}\), \n\({5 \choose 2} x^{3}(-1)^2 = 10*x^{3}*1 = 10x^{3}\), \n\({5 \choose 3} x^{2}(-1)^3 = 10*x^{2}*(-1) = -10x^{2}\), \n\({5 \choose 4} x^{1}(-1)^4 = 5*x*1 = 5x\), \n\({5 \choose 5} x^{0}(-1)^5 = 1*1*(-1) = -1\). We combine all terms to get the expanded form of the binomial.
05

Combine All Terms

Combine all terms to get the final answer: \((x-1)^{5} = x^{5} - 5x^{4} + 10x^{3} - 10x^{2} + 5x - 1\)

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Most popular questions from this chapter

Use the formula for the value of an annuity to solve Exercises. Round answers to the nearest dollar. To offer scholarship funds to children of employees, a company invests \(\$ 10,000\) at the end of every three months in an annuity that pays \(10.5 \%\) compounded quarterly. a. How much will the company have in scholarship funds at the end of ten years? b. Find the interest.

A deposit of 6000 dollars is made in an account that earns \(6 \%\) interest compounded quarterly. The balance in the account after \(n\) quarters is given by the sequence $$a_{n}=6000\left(1+\frac{0.06}{4}\right)^{n}, \quad n=1,2,3, \ldots$$ Find the balance in the account after five years. Round to the nearest cent.

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Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The sum of the geometric series \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{512}\) can only be estimated without knowing precisely what terms occur between \(\frac{1}{8}\) and \(\frac{1}{512}\).
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