Solved: "Value of ∑n=0194712n+21947?"

SnuluddidaPalbx4

Answered question

2022-02-26

Value of

\sum_{n = 0}^{1947} \frac{1}{2^{n} + \sqrt{2^{1947}}}

nastaja1en

Beginner2022-02-27Added 2 answers

We will use the property that

\sum_{r = a}^{b} f (r) = \sum_{r = a}^{b} f (a + b - r)

Therefore

S = \sum_{n = 0}^{1947} \frac{1}{2^{n} + \sqrt{2^{1947}}} = \sum_{n = 0}^{1947} \frac{1}{2^{1947 - n} + \sqrt{2^{1947}}}

Let

k = 1947

S = \sum_{n = 0}^{k} \frac{1}{2^{n} + \sqrt{2^{k}}} = \sum_{n = 0}^{k} \frac{1}{2^{k - n} + \sqrt{2^{k}}}

Now we use second expression

S = \sum_{n = 0}^{k} \frac{1}{2^{k - n} + \sqrt{2^{k}}}

S = \sum_{n = 0}^{k} \frac{2^{n}}{2^{k} + 2^{n} \sqrt{2^{k}}}

S = \frac{1}{\sqrt{2^{k}}} \sum_{n = 0}^{k} \frac{2^{n}}{\sqrt{2^{k}} + 2^{n}}

S = \frac{1}{\sqrt{2^{k}}} (\sum_{n = 0}^{k} \frac{2^{n} + \sqrt{2^{k}}}{\sqrt{2^{k}} + 2^{n}} - \sum_{n = 0}^{k} \frac{\sqrt{2^{k}}}{\sqrt{2^{k}} + 2^{n}})

S = \frac{1}{\sqrt{2^{k}}} ((k + 1) - \sqrt{2^{k}} S)

2 S = \frac{k + 1}{\sqrt{2^{k}}}

2 S = \frac{1948}{\sqrt{2^{1947}}}

S = \frac{974}{\sqrt{2^{1947}}}

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