arabella wrote:

\(2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} =\)

A \(2^{22}(2^{(23)}-1)\)

B \(2^{22}(2^{(24)}-1)\)

C \(2^{22}(2^{(25)}-1)\)

D \(2^{23}(2^{(21)}-1)\)

E \(2^{23}(2^{(22)}-1)\)

Let’s begin by factoring out 2^22, which is common to each of the terms:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22)

To find an expression for 1 + 2 + 2^2 + … + 2^22, we will use the formula

a^n - 1 = (a - 1)(a^(n - 1) + a^(n - 2) + … + a + 1).

If we let a = 2 and n = 23, we obtain

2^23 - 1 = (2 - 1)(2^22 + 2^21 + … + 2 + 1).

Thus:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22) = 2^22(2^23 - 1)

Answer: A

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