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k-1-n-5-1-k-1-5-1-k-k-1-




Question Number 140749 by SOMEDAVONG last updated on 12/May/21
Σ_(k=1) ^n 5^(1/k) (1−5^(−(1/(k(k+1)))) )=?
$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{k}}} \left(\mathrm{1}−\mathrm{5}^{−\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}} \right)=? \\ $$
Answered by Dwaipayan Shikari last updated on 12/May/21
5^(1/k) (1−5^(−(1/k)+(1/(k+1))) )=5^(1/k) −5^(1/(k+1))   Σ_(k=1) ^n (5^(1/k) −5^(1/(k+1)) )=5−(√5)+(√5)−(5)^(1/3) +..+5^(1/n) −5^(1/(n+1)) =5−5^(1/(n+1))
$$\mathrm{5}^{\frac{\mathrm{1}}{{k}}} \left(\mathrm{1}−\mathrm{5}^{−\frac{\mathrm{1}}{{k}}+\frac{\mathrm{1}}{{k}+\mathrm{1}}} \right)=\mathrm{5}^{\frac{\mathrm{1}}{{k}}} −\mathrm{5}^{\frac{\mathrm{1}}{{k}+\mathrm{1}}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{5}^{\frac{\mathrm{1}}{{k}}} −\mathrm{5}^{\frac{\mathrm{1}}{{k}+\mathrm{1}}} \right)=\mathrm{5}−\sqrt{\mathrm{5}}+\sqrt{\mathrm{5}}−\sqrt[{\mathrm{3}}]{\mathrm{5}}+..+\mathrm{5}^{\frac{\mathrm{1}}{{n}}} −\mathrm{5}^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} =\mathrm{5}−\mathrm{5}^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} \\ $$

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