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given-N-4-5-1-5-1-4-1-5-1-8-1-5-1-16-1-find-value-of-N-1-48-

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Question Number 76438 by john santu last updated on 27/Dec/19
given N =(4/(((√5)+1)(5^(1/4) +1)(5^(1/8) +1)(5^(1/(16)) +1))) find value of (N+1)^(48) .
$${given}\:{N}\:=\frac{\mathrm{4}}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)} \\ $$$${find}\:{value}\:{of}\:\left({N}+\mathrm{1}\right)^{\mathrm{48}} . \\ $$
Answered by mr W last updated on 27/Dec/19
N =(4/(((√5)+1)(5^(1/4) +1)(5^(1/8) +1)(5^(1/(16)) +1))) N =((4(5^(1/(16)) −1))/(((√5)+1)(5^(1/4) +1)(5^(1/8) +1)(5^(1/(16)) +1)(5^(1/(16)) −1))) N =((4(5^(1/(16)) −1))/(((√5)+1)(5^(1/4) +1)(5^(1/8) +1)(5^(1/8) −1))) N =((4(5^(1/(16)) −1))/(((√5)+1)(5^(1/4) +1)(5^(1/4) −1))) N =((4(5^(1/(16)) −1))/(((√5)+1)((√5)−1))) N =((4(5^(1/(16)) −1))/((5−1))) N =5^(1/(16)) −1 N+1 =5^(1/(16)) (N+1)^(48) =5^((48)/(16)) (N+1)^(48) =5^3 (N+1)^(48) =125
$${N}\:=\frac{\mathrm{4}}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)} \\ $$$${N}\:=\frac{\mathrm{4}\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right)} \\ $$$${N}\:=\frac{\mathrm{4}\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{8}}} −\mathrm{1}\right)} \\ $$$${N}\:=\frac{\mathrm{4}\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}\right)} \\ $$$${N}\:=\frac{\mathrm{4}\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)} \\ $$$${N}\:=\frac{\mathrm{4}\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right)}{\left(\mathrm{5}−\mathrm{1}\right)} \\ $$$${N}\:=\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1} \\ $$$${N}+\mathrm{1}\:=\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$\left({N}+\mathrm{1}\right)^{\mathrm{48}} \:=\mathrm{5}^{\frac{\mathrm{48}}{\mathrm{16}}} \\ $$$$\left({N}+\mathrm{1}\right)^{\mathrm{48}} \:=\mathrm{5}^{\mathrm{3}} \\ $$$$\left({N}+\mathrm{1}\right)^{\mathrm{48}} \:=\mathrm{125} \\ $$
Commented by john santu last updated on 28/Dec/19
thanks you. gbu
$${thanks}\:{you}.\:{gbu}\: \\ $$

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