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Algebra Difficulty 4.8 AIME Find the answer

Find the sum 2141−1+2242−1+2444−1+2848−1+⋯\frac{2^{1}}{4^{1}-1}+\frac{2^{2}}{4^{2}-1}+\frac{2^{4}}{4^{4}-1}+\frac{2^{8}}{4^{8}-1}+\cdots

A number or a short expression. Spacing and $ signs are ignored.

Solution

Notice that 22k42k−1=22k+142k−1−142k−1=122k−1−142k−1=142k−1−1−142k−1\frac{2^{2^{k}}}{4^{2^{k}}-1}=\frac{2^{2^{k}}+1}{4^{2^{k}}-1}-\frac{1}{4^{2^{k}}-1}=\frac{1}{2^{2^{k}}-1}-\frac{1}{4^{2^{k}}-1}=\frac{1}{4^{2^{k-1}}-1}-\frac{1}{4^{2^{k}}-1} Therefore, the sum telescopes as (142−1−1−1420−1)+(1420−1−1421−1)+(1421−1−1422−1)+⋯\left(\frac{1}{4^{2^{-1}}-1}-\frac{1}{4^{2^{0}}-1}\right)+\left(\frac{1}{4^{2^{0}}-1}-\frac{1}{4^{2^{1}}-1}\right)+\left(\frac{1}{4^{2^{1}}-1}-\frac{1}{4^{2^{2}}-1}\right)+\cdots and evaluates to 1/(42−1−1)=11 /\left(4^{2^{-1}}-1\right)=1.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.