How do I find the sum:


The method of telescoping surely won't work for this, and I tried rationalizing each terms to try and cancel consecutive terms but it didnt work for me. How do I go about approaching this?

  • $\begingroup$ Hint: if $a_n=1/(2^n+\sqrt{2^{1947}})$, then $a_n+a_{1947-n}=\ldots$ $\endgroup$
    – metamorphy
    Oct 20, 2019 at 19:17
  • $\begingroup$ I can't seem to be able to simplify $a_0+a_{1947}...$ could you help? Also, how do I notice these kinds of patterns while solving questions? $\endgroup$
    – Techie5879
    Oct 22, 2019 at 4:32
  • $\begingroup$ I’ve edited the title and the question modifying the term series by summation. $\endgroup$
    – user
    Sep 5, 2021 at 15:25

1 Answer 1


We have

$$\sum_{n=0}^{1947} \frac{1}{ 2^n+\sqrt{2^{1947}} }=\frac1{\sqrt{2^{1947}} }\sum_{n=0}^{1947} \frac{1}{2^{n-\frac{1947}2}+1}=\frac1{\sqrt{2^{1947}}}\sum_{n=0}^{1947}a_n$$



and so on for the remaining terms.


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