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Question Number 155465 by cortano last updated on 01/Oct/21
(1^2 /(1×3))+(2^2 /(3×5))+(3^2 /(5×7))+...+(n^2 /((2n−1)(2n+1)))=
$$\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}×\mathrm{5}}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{5}×\mathrm{7}}+…+\frac{\mathrm{n}^{\mathrm{2}} }{\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}= \\ $$
Answered by Ar Brandon last updated on 03/Oct/21
S=Σ_(k=1) ^n (k^2 /((2k−1)(2k+1)))=(1/(16))Σ_(k=1) ^n ((((2k−1)+(2k+1))^2 )/((2k−1)(2k+1))) =(1/(16))Σ_(k=1) ^n (((2k−1)/(2k+1))+2+((2k+1)/(2k−1)))=(1/(16))Σ_(k=1) ^n (1−(2/(2k+1))+2+1+(2/(2k−1))) =(1/(16))Σ_(k=1) ^n (4+(2/(2k−1))−(2/(2k+1)))=(n/4)+(1/8)Σ_(k=1) ^n ((1/(2k−1))−(1/(2k+1))) =(n/4)+(1/8)(1−(1/3)+(1/3)−(1/5)+∙∙∙+(1/(2n−3))−(1/(2n−1))+(1/(2n−1))−(1/(2n+1))) =(n/4)+(1/8)(1−(1/(2n+1)))=(n/4)+(n/(4(2n+1)))=((2n^2 +2n)/(4(2n+1)))=((n^2 +n)/(4n+2))
$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{16}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(\left(\mathrm{2}{k}−\mathrm{1}\right)+\left(\mathrm{2}{k}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}+\mathrm{2}+\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{16}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}{k}+\mathrm{1}}+\mathrm{2}+\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}{k}−\mathrm{1}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{4}+\frac{\mathrm{2}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{2}}{\mathrm{2}{k}+\mathrm{1}}\right)=\frac{{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right) \\ $$$$\:\:\:\:=\frac{{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$\:\:\:\:=\frac{{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)=\frac{{n}}{\mathrm{4}}+\frac{{n}}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{4}{n}+\mathrm{2}} \\ $$

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