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Chapter 15: Problem 29

How many terms are in the expansion of \((a+b)^{9} ?\)

Short Answer

Expert verified
There are 10 terms in the expansion of \((a+b)^{9}\).

Step by step solution

01

Recall the binomial theorem

The binomial theorem states that for any natural number n, the expansion of \((a+b)^n\) can be written as a sum of terms, each of which has a "binomial coefficient" multiplied by a power of a and a power of b. In general, the binomial theorem can be written as: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\), where \(\binom{n}{k}\) denotes the binomial coefficient, which is defined as: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where n! denotes the factorial of n (i.e., the product of all positive integers up to and including n).
02

Determine the number of terms using the binomial theorem

In our case, we have \(n = 9\), so the expansion of \((a+b)^9\) will have a sum of terms from \(k = 0\) to \(k = n\). This means that the expansion will have \(n + 1\) terms, because we start counting from \(k = 0\). So, the number of terms in the expansion of \((a+b)^9\) will be: \(9 + 1 = 10\).
03

Final answer

There are 10 terms in the expansion of \((a+b)^{9}\).

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Most popular questions from this chapter

In your own words, explain how to evaluate \(n !\) for any positive integer.

Use Pascal鈥檚 Triangle to expand each binomial. $$(y+z)^{5}$$

Use the formula for \(S_{n}\) to find the sum of the terms of each geometric sequence. $$\sum_{i=1}^{5} 2\left(\frac{1}{3}\right)^{i}$$

Evaluate each binomial coefficient. $$\left(\begin{array}{c}11 \\\8\end{array}\right)$$

Use the formula for \(S_{n}\) to find the sum of the terms of each geometric sequence. $$-\frac{1}{4},-\frac{1}{2},-1,-2,-4,-8$$
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