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Chapter 6: Problem 33

Express each of the sums in \(24-35\) in closed form (without using a summation symbol and without using an ellipsis \(\cdots\) ). $$ \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{l} n \\ k \end{array}\right) 3^{2 n-2 k} 2^{2 k} $$

Short Answer

Expert verified
The closed form for the given sum is \(\sqrt{13}^n \cos(n\arctan\frac{2}{3})\).

Step by step solution

01

Understanding the structure of the sum

The sum goes from k=0 to n. For each k, the general term is made up of three parts: \((-1)^k\), \(\binom{n}{k}\), and \(3^{2n-2k} \cdot 2^{2k}\). Here, \((-1)^k\) creates an alternating sign, \(\binom{n}{k}\) is a binomial coefficient, and \(3^{2n-2k} \cdot 2^{2k} = (3^{n-k} \cdot 2^k)^2\), which recombines to get a square of a base-6 term.
02

Express the General Term Differently

We want the sequence of terms to look more natural, so we express the general term differently. Notice that \((3^{n-k} \cdot 2^k)^2\) is the square of \(6^k\). Also using an identity of the binomial theorem \((a+b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^k\), let's let \(a = 3\) and \(b = 2i\). Then, we find the term \((-1)^k\binom{n}{k}(3^{n-k}(2i)^k)= (-1)^k\binom{n}{k}((3+2i)^n + (3-2i)^n)\) is represented.
03

Simplify the expression

Now, when we look at our original sequence \(\sum_{k=0}^{n}(-1)^{k} \binom{n}{k} 3^{2n-2k} 2^{2k}\), we recognize it is similar with what we get in step 2 with an exception. The binomial theorem doesn't come with an alternating sign. As an ingenius idea, we remember that \(\Re((3+2i)^n) = \Re((3+2i)^n + (3-2i)^n)\). Therefore, we conclude that the expression equals to the real part of \((3+2i)^n\), because the alternating sign essentially selects the real part of the complex number.
04

Find the closed form for the sum

We have shown that the sum equals to the real part of the complex number \((3+2i)^n = |3+2i|^n (\cos(n\varphi) + i\sin(n\varphi))\), where \(|3+2i| = \sqrt{3^2+2^2} = \sqrt{13}\) is the modulus and \(\varphi = \arctan(\frac{2}{3})\) is the argument of the complex number \(3+2i\). Therefore, the closed form for the sum is that it equals \(\sqrt{13}^n \cos(n\arctan\frac{2}{3})\).

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