By induction, it is pretty clear that x2n+1−1=(x−1)∏k=0n(x2k+1) and by applying x⋅ddxlog⁡(.) to both sides we have: 2n+1x2n+1x2n+1−1=xx−1+∑k=0n2kx2kx2k+1 and by replacing x with 1z: 2n+11−z2n+1=11−z+∑k=0n2kz2k+1

Hint: If f(x)=11+x+21+x2+41+x4+…+ up to n+1 terms 11−x+f(x)=11−x+11+x+(21+x2+41+x4+…+ up to n+1 terms) =21−x2+21+x2+(41+x4+…+ up to n terms) =41−x4+41+x4+(81+x8+…+ up to n−1 terms)

We see an example of telescoping based upon 2k1−x2k+2k1+x2k=2k(1+x2k)+2k(1−x2k)(1−x2k)(1+x2k)=2k+11−x2k+1 We obtain according to (1) ∑k=0n2k1+x2k=∑k=0n(2k+11−x2k+1−2k1−x2k) =∑k=0n2k+11−x2k+1−∑k=0n2k1−x2k =∑k=1n+12k1−x2k−∑k=0n2k1−x2k =2n+11−x2n+1−11−x

By induction, it is pretty clear that x2n+1−1=(x−1)∏k=0n(x2k+1) and by applying x⋅ddxlog⁡(.) to both sides we have: 2n+1x2n+1x2n+1−1=xx−1+∑k=0n2kx2kx2k+1 and by replacing x with 1z: 2n+11−z2n+1=11−z+∑k=0n2kz2k+1

Hint: If f(x)=11+x+21+x2+41+x4+…+ up to n+1 terms 11−x+f(x)=11−x+11+x+(21+x2+41+x4+…+ up to n+1 terms) =21−x2+21+x2+(41+x4+…+ up to n terms) =41−x4+41+x4+(81+x8+…+ up to n−1 terms)

We see an example of telescoping based upon 2k1−x2k+2k1+x2k=2k(1+x2k)+2k(1−x2k)(1−x2k)(1+x2k)=2k+11−x2k+1 We obtain according to (1) ∑k=0n2k1+x2k=∑k=0n(2k+11−x2k+1−2k1−x2k) =∑k=0n2k+11−x2k+1−∑k=0n2k1−x2k =∑k=1n+12k1−x2k−∑k=0n2k1−x2k =2n+11−x2n+1−11−x

"Theres" solved and explained

Branden Valentine

Answered question

2022-01-22

Theres

Answer & Explanation

hmotans

Beginner2022-01-23Added 8 answers

By induction, it is pretty clear that

x^{2^{n + 1}} - 1 = (x - 1) \prod_{k = 0}^{n} (x^{2^{k}} + 1)

and by applying

x \cdot \frac{d}{d x} \log (.)

to both sides we have:

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

and by replacing x with

\frac{1}{z}

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

Joy Compton

Beginner2022-01-24Added 13 answers

Hint:
If

f (x) = \frac{1}{1 + x} + \frac{2}{1 + x^{2}} + \frac{4}{1 + x^{4}} + \dots +

up to

n + 1

terms

\frac{1}{1 - x} + f (x) = \frac{1}{1 - x} + \frac{1}{1 + x} + (\frac{2}{1 + x^{2}} + \frac{4}{1 + x^{4}} + \dots + up to n + 1 terms)

= \frac{2}{1 - x^{2}} + \frac{2}{1 + x^{2}} + (\frac{4}{1 + x^{4}} + \dots + up to n terms)

= \frac{4}{1 - x^{4}} + \frac{4}{1 + x^{4}} + (\frac{8}{1 + x^{8}} + \dots + up to n - 1 terms)

RizerMix

Expert2022-01-27Added 656 answers

We see an example of telescoping based upon

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

We obtain according to (1)

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

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

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

= \frac{2^{n + 1}}{1 - x^{2^{n + 1}}} - \frac{1}{1 - x}

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