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Question Number 197838 by hardmath last updated on 30/Sep/23

Answered by MM42 last updated on 30/Sep/23

$$\frac{2^{49}}{2^{49}+1}+\frac{2^{48}}{2^{48}+1}+...+\frac{1}{2^{48}+1}+\frac{1}{2^{49}+1}$$$$=1+1+...+1=49✓$$

Commented by hardmath last updated on 30/Sep/23

$$\mathrm{Is}\mathrm{this}\mathrm{the}\mathrm{right}\mathrm{solution},\mathrm{professor}?$$

Commented by cortano12 last updated on 01/Oct/23

$$49\frac{1}{2}$$

Commented by Rasheed.Sindhi last updated on 01/Oct/23

$$(\frac{2^{49}}{2^{49}+1}+\frac{1}{2^{49}+1})+(\frac{2^{48}}{2^{48}+1}+\frac{1}{2^{48}+1})+...+(\frac{2^2}{2^2+1}+\frac{1}{2^2+1})+(\frac{2^1}{2^1+1}+\frac{1}{2^1+1})+\left(\frac{1}{2^0+1}\right)$$$$\frac{2^{49}+1}{2^{49}+1}+\frac{2^{48}+1}{2^{48}+1}+...+\frac{2^1+1}{2^1+1}+\frac{1}{1+1}$$$$1+1+1+...49timed+\frac{1}{2}$$$$=49\frac{1}{2}$$

Commented by MM42 last updated on 01/Oct/23

$$ok.✓$$

Answered by Rasheed.Sindhi last updated on 02/Oct/23

$$\mathrm{AnOther}\mathrm{way}...$$$$\frac{1}{2^{-49}+1}+\frac{1}{2^{-48}+1}+...+\frac{1}{2^{48}+1}+\frac{1}{2^{49}+1}$$$$=(\frac{1}{2^{-49}+1}+\frac{1}{2^{49}+1})+(\frac{1}{2^{-48}+1}+\frac{1}{2^{48}+1})+$$$$....+(\frac{1}{2^{-1}+1}+\frac{1}{2^1+1})+\left(\frac{1}{2^0+1}\right)$$$$=\left(\frac{2^{49}+1+2^{-49}+1}{1+2^{-49}+2^{49}+1}\right)+\left(\frac{2^{48}+1+2^{-48}+1}{1+2^{-48}+2^{48}+1}\right)+$$$$...+\left(\frac{2^1+1+2^{-1}+1}{1+2^{-1}+2^1+1}\right)+\left(\frac{1}{1+1}\right)$$$$=1+1+....+1\left(49times\right)+\frac{1}{2}$$$$=49\frac{1}{2}$$$$$$

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