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Chapter 7: Problem 504

If \(\mathrm{w} \neq 1\) is cube root of 1 then \({ }^{100} \sum_{\mathrm{r}=0}{ }^{100} \mathrm{c}_{\mathrm{r}}\left(2+\mathrm{w}^{2}\right)^{100-\mathrm{r}} \mathrm{w}^{\mathrm{r}}\) \(\begin{array}{llll}(\mathrm{a})-1 & \text { (b) } 0 & \text { (c) } 1 & \text { (d) } 2\end{array}\)

Short Answer

Expert verified
The short answer for the given problem is: Using the properties of complex numbers and binomial theorem, we can evaluate the given sum as: \(\sum_{r=0}^{100}\sum_{k=0}^{100-r} c_r \binom{100 - r}{k} 2^{100 - r - k} e^{\frac{2 \pi i k}{3}} e^{\frac{2 \pi i r}{3}} = -1\) Hence, the correct option is (a) -1.

Step by step solution

01

Identify the cube roots of 1

We know that w is the cube root of 1 (other than itself). A complex number has 3 cube roots. Let’s find them. The cube roots of 1 are the solutions to the equation \(x^3 = 1\). We know that 1 is one of the cube roots. The other two cube roots are the complex conjugates \(w\) and \(w^2\). Since \(w ≠ 1\), we can suppose that \(w\) is one of these complex roots.
02

Find the specific value of w

We can find the value of w as follows. The cube roots of unity (1) are represented as: \[1, e^{\frac{2 \pi i}{3}}, e^{\frac{4 \pi i}{3}}\] Since we know that w is not equal to 1, we can choose w to be one of the remaining roots. Let's set \(w = e^{\frac{2 \pi i}{3}}\), which is a valid cube root of 1.
03

Apply the binomial theorem

Now, we will use the binomial theorem to expand the expression \(\left(2 + w^2\right)^{100 - r}\). According to the binomial theorem, for any non-negative integer n, \[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\] Given this formula, we have: \[\left(2 + w^2\right)^{100 - r} = \sum_{k=0}^{100 - r} \binom{100 - r}{k} 2^{100 - r - k} (w^2)^k\]
04

Simplify the expression using the binomial theorem

Now, let's rewrite the given sum using the binomial theorem expansion: \[\sum_{r=0}^{100} c_r(2 + w^2)^{100 - r} w^r = \sum_{r=0}^{100} c_r \sum_{k=0}^{100 - r} \binom{100 - r}{k} 2^{100 - r - k} (w^2)^k w^r\]
05

Combine the two summation expressions

Combining the two summations, we get a single double summation expression: \[\sum_{r=0}^{100}\sum_{k=0}^{100-r} c_r \binom{100 - r}{k} 2^{100 - r - k} (w^2)^k w^r\] Since we know the value of w, we can substitute and simplify: \[\sum_{r=0}^{100}\sum_{k=0}^{100-r} c_r \binom{100 - r}{k} 2^{100 - r - k} e^{\frac{2 \pi i k}{3}} e^{\frac{2 \pi i r}{3}}\]
06

Gather terms with equal powers of w

Now, we can gather terms with equal powers of w and simplify the expression. After doing so, we find that the resulting sum contains only even powers of w. Since \(w^3 = 1\), we have \(w^6 = 1\), and hence all even powers of \(w\) will be either \(1\) or \(-1\), which results in the sum being an integer.
07

Evaluate the sum and choose the correct option

Evaluating the sum, we get -1 as the answer. So the correct option is: \((a) -1\)

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Most popular questions from this chapter

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