Question Number 159891 by tounghoungko last updated on 22/Nov/21
$$\:\:\:\:{S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2002}\:} {\sum}}\sqrt{\frac{{k}^{\mathrm{2}} +\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }}\:=? \\ $$
Answered by chhaythean last updated on 22/Nov/21
![S=Σ_(k=1) ^(2002) (√(((k^2 +1)/k^2 )+(1/((k+1)^2 )))) Notice that: ((k^2 +1)/k^2 )+(1/((k+1)^2 )) =(((k^2 +1)(k+1)^2 +k^2 )/(k^2 (k+1)^2 )) =((k^4 +2k^3 +k^2 +k^2 +2k+1+k^2 )/(k^2 (k+1)^2 )) =((k^2 (k+1)^2 +2k(k+1)+1)/(k^2 (k+1)^2 )) =(([k(k+1)+1]^2 )/(k^2 (k+1)^2 )) ⇒(√(((k^2 +1)/k^2 )+(1/((k+1)^2 ))))=((k(k+1)+1)/(k(k+1)))=1+(1/k)−(1/((k+1))) therefore: S=Σ_(k=1) ^(2002) 1+Σ_(k=1) ^(2002) ((1/k)−(1/(k+1))) =2002+1−(1/(2003)) So determinant (((S=((4012008)/(2003)))))](https://www.tinkutara.com/question/Q159901.png)
$$\mathrm{S}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2002}} {\sum}}\sqrt{\frac{\mathrm{k}^{\mathrm{2}} +\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{Notice}\:\mathrm{that}:\:\frac{\mathrm{k}^{\mathrm{2}} +\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{k}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} }{\mathrm{k}^{\mathrm{2}} \left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{k}^{\mathrm{4}} +\mathrm{2k}^{\mathrm{3}} +\mathrm{k}^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} +\mathrm{2k}+\mathrm{1}+\mathrm{k}^{\mathrm{2}} }{\mathrm{k}^{\mathrm{2}} \left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{k}^{\mathrm{2}} \left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2k}\left(\mathrm{k}+\mathrm{1}\right)+\mathrm{1}}{\mathrm{k}^{\mathrm{2}} \left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left[\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)+\mathrm{1}\right]^{\mathrm{2}} }{\mathrm{k}^{\mathrm{2}} \left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{\frac{\mathrm{k}^{\mathrm{2}} +\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)+\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)} \\ $$$$\mathrm{therefore}:\:\mathrm{S}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2002}} {\sum}}\mathrm{1}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2002}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2002}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2003}} \\ $$$$\mathrm{So}\:\begin{array}{|c|}{\mathrm{S}=\frac{\mathrm{4012008}}{\mathrm{2003}}}\\\hline\end{array} \\ $$
Commented by tounghoungko last updated on 22/Nov/21
$${yes}\:{telescopic}\:{series} \\ $$