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Question Number 118880 by bemath last updated on 20/Oct/20

$$8(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})(1+\frac{1}{2^8})...(1+\frac{1}{2^{32}})+\frac{1}{2^{60}}=?$$

Answered by Dwaipayan Shikari last updated on 20/Oct/20

$$8(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3})(1+\frac{1}{2^4}+\frac{1}{2^8}+\frac{1}{2^{12}})(1+\frac{1}{2^{16}})(1+\frac{1}{2^{32}})+\frac{1}{2^{60}}$$$$8\left(\frac{1-\frac{1}{2^4}}{\frac{1}{2}}\right)\left(\frac{1-\frac{1}{2^{16}}}{1-\frac{1}{2^4}}\right)\left(\frac{1-\frac{1}{2^{64}}}{1-\frac{1}{2^{16}}}\right)+\frac{1}{2^{60}}$$$$=63\left(\frac{2^{16}-1}{2^{12}(2^4-1)}\right)\left(\frac{2^{64}-1}{2^{48}(2^{16}-1)}\right)+\frac{1}{2^{60}}$$$$=\frac{2^{64}-1}{2^{60}}+\frac{1}{2^{60}}=2^4=16$$

Answered by bobhans last updated on 20/Oct/20

$$let8(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}})+\frac{1}{2^{60}}=z$$$$multiplyby(1-\frac{1}{2})bothsidesgives$$$$8(1-\frac{1}{2})(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}})+\frac{1}{2^{61}}=\frac{1}{2}z$$$$\Rightarrow 8(1-\frac{1}{2^2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{32})+\frac{1}{2^{61}}=\frac{1}{2}z$$$$\Rightarrow 8(1-\frac{1}{2^4})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}})+\frac{1}{2^{61}}=\frac{1}{2}z$$$$\Rightarrow 8(1-\frac{1}{2^{64}})+\frac{1}{2^{61}}=\frac{1}{2}z$$$$\Rightarrow 8\left(\frac{2^{64}-1}{2^{64}}\right)+\frac{1}{2^{61}}=\frac{1}{2}z$$$$\Rightarrow \frac{2^{64}-1}{2^{61}}+\frac{1}{2^{61}}=\frac{1}{2}z$$$$\Rightarrow \frac{2^{64}}{2^{61}}=\frac{1}{2}z,hencez=2\times 2^3=16$$

Answered by 1549442205PVT last updated on 21/Oct/20

$$\mathrm{Apply}(\mathrm{a}-\mathrm{b}(\mathrm{a}+\mathrm{b})={\mathrm{a}}^2-{\mathrm{b}}^2\mathrm{we}\mathrm{have}$$$$\mathrm{P}=8(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})(1+\frac{1}{2^8})...(1+\frac{1}{2^{32}})+\frac{1}{2^{61}}$$$$\iff (1-\frac{1}{2})\mathrm{P}=8(1-\frac{1}{2^2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})(1+\frac{1}{2^8})...(1+\frac{1}{2^{32}})+\frac{1}{2^{61}}$$$$=8(1-\frac{1}{2^4})(1+\frac{1}{2^4})(1+\frac{1}{2^8})...(1+\frac{1}{2^{32}})+\frac{1}{2^{61}}$$$$=8(1-\frac{1}{2^8})(1+\frac{1}{2^8})...(1+\frac{1}{2^{32}})+\frac{1}{2^{61}}=...$$$$=8(1-\frac{1}{2^{32}})(1+\frac{1}{2^{32}})+\frac{1}{2^{61}}=8(1-\frac{1}{2^{64}})+\frac{1}{2^{61}}$$$$=8-\frac{8}{2^{64}}+\frac{1}{2^{61}}=8\Rightarrow \mathrm{P}=8.2=16$$

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