Problem 75
Question
Write the power series for \((1+x)^{k}\) in terms of binomial coefficients.
Step-by-Step Solution
Verified Answer
\[(1+x)^{k} = \binom{k}{0}x^0 + \binom{k}{1}x^1 + \binom{k}{2}x^2 +...+ \binom{k}{k}x^k\]
1Step 1: Declaring the Binomial Theorem
The Binomial theorem is a mathematical formula that describes the algebraic expansion of powers of a binomial. It is represented as: \[(1+x)^{k} = \sum_{n=0}^{k} \binom{k}{n} x^n\] where \(k\) is any real number and \(\binom{k}{n}\) represents the binomial coefficient.
2Step 2: Understanding Binomial Coefficients
The binomial coefficient, \(\binom{k}{n}\), is the number of ways in which you can choose \(n\) elements from a pool of \(k\) elements. It can be computed using the formula: \[\binom{k}{n} = \frac{k!}{n!(k-n)!}\] where \(k!\) denotes the factorial of \(k\), which means the product of all positive integers less than or equal to \(k\).
3Step 3: Applying the Binomial Theorem
By the Binomial theorem, the power series for \((1+x)^{k}\) in terms of binomial coefficients is: \[(1+x)^{k} = \binom{k}{0}x^0 + \binom{k}{1}x^1 + \binom{k}{2}x^2 +...+ \binom{k}{k}x^k\] Each term of the power series corresponds to a binomial coefficient. The power on \(x\) in each term also corresponds to the number of ways to choose elements from a pool of \(k\) elements.
Other exercises in this chapter
Problem 75
(a) use Theorem 7.5 to show that the sequence with the given \(n\) th term converges and (b) use a graphing utility to graph the first 10 terms of the sequence
View solution Problem 75
Use the divergence test given in Exercise 71 to show that the series diverges. $$ \sum_{n=1}^{\infty} \frac{n^{3}}{5 n^{4}+3} $$
View solution Problem 75
In Exercises \(75-88\), determine the convergence or divergence of the series using any appropriate test from this chapter. Identify the test used. \(\sum_{n=1}
View solution Problem 76
(a) use Theorem 7.5 to show that the sequence with the given \(n\) th term converges and (b) use a graphing utility to graph the first 10 terms of the sequence
View solution