Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 155465 by cortano last updated on 01/Oct/21

$$\frac{1^2}{1\times 3}+\frac{2^2}{3\times 5}+\frac{3^2}{5\times 7}+...+\frac{{\mathrm{n}}^2}{(2\mathrm{n}-1)(2\mathrm{n}+1)}=$$

Answered by Ar Brandon last updated on 03/Oct/21

$$S=\underset{k=1}{\sum ^n}\frac{k^2}{(2k-1)(2k+1)}=\frac{1}{16}\underset{k=1}{\sum ^n}\frac{{((2k-1)+(2k+1))}^2}{(2k-1)(2k+1)}$$$$=\frac{1}{16}\underset{k=1}{\sum ^n}(\frac{2k-1}{2k+1}+2+\frac{2k+1}{2k-1})=\frac{1}{16}\underset{k=1}{\sum ^n}(1-\frac{2}{2k+1}+2+1+\frac{2}{2k-1})$$$$=\frac{1}{16}\underset{k=1}{\sum ^n}(4+\frac{2}{2k-1}-\frac{2}{2k+1})=\frac{n}{4}+\frac{1}{8}\underset{k=1}{\sum ^n}(\frac{1}{2k-1}-\frac{1}{2k+1})$$$$=\frac{n}{4}+\frac{1}{8}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdot \cdot \cdot +\frac{1}{2n-3}-\frac{1}{2n-1}+\frac{1}{2n-1}-\frac{1}{2n+1})$$$$=\frac{n}{4}+\frac{1}{8}(1-\frac{1}{2n+1})=\frac{n}{4}+\frac{n}{4(2n+1)}=\frac{2n^2+2n}{4(2n+1)}=\frac{n^2+n}{4n+2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com