Question-197194

Question Number 197194 by sonukgindia last updated on 10/Sep/23

Answered by witcher3 last updated on 10/Sep/23

k^(1−n) +(k+1)^(1−n) =(((1+k)^(n−1) +k^(n−1) )/((k(1+k))^(n−1) )) ⇔Σ_(k=1) ^(50) (((k^(n−1) +(1+k)^(n−1) )/((k(k+1))^(n−1) (k^(n−1) +(1+k)^(n−1) ))))^(1/(n−1)) =Σ_(k=1) ^(50) (1/(k(k+1)))=Σ_(k=1) ^(50) (1/k)−(1/(k+1))=1−(1/(51))=((50)/(51))

$$\mathrm{k}^{\mathrm{1}−\mathrm{n}} +\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{1}−\mathrm{n}} =\frac{\left(\mathrm{1}+\mathrm{k}\right)^{\mathrm{n}−\mathrm{1}} +\mathrm{k}^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{k}\left(\mathrm{1}+\mathrm{k}\right)\right)^{\mathrm{n}−\mathrm{1}} } \\ $$$$\Leftrightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{50}} {\sum}}\sqrt[{\mathrm{n}−\mathrm{1}}]{\frac{\mathrm{k}^{\mathrm{n}−\mathrm{1}} +\left(\mathrm{1}+\mathrm{k}\right)^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{k}^{\mathrm{n}−\mathrm{1}} +\left(\mathrm{1}+\mathrm{k}\right)^{\mathrm{n}−\mathrm{1}} \right)}} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{50}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{50}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{51}}=\frac{\mathrm{50}}{\mathrm{51}} \\ $$

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