Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 106694 by 175mohamed last updated on 06/Aug/20

Answered by bobhans last updated on 06/Aug/20

$${\mathrm{U}}_{\mathrm{n}}=\frac{2\mathrm{n}+1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}\Rightarrow \underset{\mathrm{n}=1}{\sum ^{40}}\frac{2\mathrm{n}+1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}$$$$\mathrm{decomposition}\mathrm{fraction}$$$$\frac{2\mathrm{n}+1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{\mathrm{A}}{\mathrm{n}}+\frac{\mathrm{B}}{\mathrm{n}+1}+\frac{\mathrm{C}}{\mathrm{n}+2}$$$$2\mathrm{n}+1=\mathrm{A}(\mathrm{n}+1)(\mathrm{n}+2)+\mathrm{Bn}(\mathrm{n}+2)+\mathrm{Cn}(\mathrm{n}+1)$$$$\mathrm{n}=0\to 1=2\mathrm{A};\mathrm{A}=\frac{1}{2}$$$$\mathrm{n}=-1\to -1=-\mathrm{B};\mathrm{B}=1$$$$\mathrm{n}=-2\to -3=2\mathrm{C};\mathrm{C}=-\frac{3}{2}$$$$\underset{\mathrm{n}=1}{\sum ^{40}}(\frac{1}{2\mathrm{n}}+\frac{1}{\mathrm{n}+1}-\frac{3}{2(\mathrm{n}+2)})$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$\underset{n=1}{\sum ^n}\frac{2n+1}{n(n+1)(n+2)}$$$$=\sum ^n\frac{n+2+n-1}{n(n+1)(n+2)}=\sum ^n\frac{1}{(n+1)(n+2)}+\sum ^n\frac{1}{n(n+1)}-\sum ^n\frac{1}{n(n+1)(n+2)}$$$$=\sum ^n\frac{1}{n+1}-\frac{1}{n+2}+\sum ^n\frac{1}{n}-\frac{1}{n+1}-\frac{1}{2}\sum ^n\frac{n+2-n}{n(n+1)(n+2)}$$$$=(\frac{1}{2}-\frac{1}{n+2})+(1-\frac{1}{n+1})-\frac{1}{2}\sum ^n\frac{1}{n(n+1)}+\frac{1}{2}\sum ^n\frac{1}{(n+1)(n+2)}$$$$=(\frac{1}{2}-\frac{1}{n+2})+(1-\frac{1}{n+1})-\frac{1}{2}\sum ^n\frac{1}{n}-\frac{1}{n+1}+\frac{1}{2}\sum ^n\frac{1}{n+1}-\frac{1}{n+2}$$$$=(\frac{1}{2}-\frac{1}{n+2})+(1-\frac{1}{n+1})-\frac{1}{2}(1-\frac{1}{n+1})+\frac{1}{2}(\frac{1}{2}-\frac{1}{n+2})$$$$=\frac{3}{2}(\frac{1}{2}-\frac{1}{n+2})+\frac{1}{2}(1-\frac{1}{n+1})$$

Answered by mathmax by abdo last updated on 06/Aug/20

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{2\mathrm{k}+1}{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}\mathrm{first}\mathrm{we}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{\mathrm{x}(\mathrm{x}+1)(\mathrm{x}+2)}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+1}+\frac{\mathrm{c}}{\mathrm{x}+2}\Rightarrow$$$$\mathrm{a}=\frac{1}{2},\mathrm{b}=\frac{-1}{-1}=1,\mathrm{c}=\frac{-3}{(-2)(-1)}=-\frac{3}{2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2\mathrm{x}}+\frac{1}{\mathrm{x}+1}-\frac{3}{2(\mathrm{x}+2)}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{k}\right)$$$$=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}+\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}-\frac{3}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}\mathrm{we}\mathrm{have}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+1}-1$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{2}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}+{\mathrm{H}}_{\mathrm{n}+1}-1-\frac{3}{2}({\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{2})$$$$=\frac{{\mathrm{H}}_{\mathrm{n}}}{2}+{\mathrm{H}}_{\mathrm{n}+1}-\frac{3}{2}{\mathrm{H}}_{\mathrm{n}+2}+\frac{5}{4}\Rightarrow$$$$=\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}+{\mathrm{H}}_{\mathrm{n}}+\frac{1}{\mathrm{n}+1}-\frac{3}{2}({\mathrm{H}}_{\mathrm{n}}+\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2})+\frac{5}{4}$$$$=\frac{1}{n+1}-\frac{3}{2(n+1)}-\frac{3}{2(n+2)}+\frac{5}{4}$$$$=-\frac{1}{2(n+1)}-\frac{3}{2(n+2)}+\frac{5}{4}\Rightarrow$$$${\mathrm{A}}_{40}=-\frac{1}{2\times 41}-\frac{3}{2\times 42}+\frac{5}{4}=\frac{5}{4}-\frac{3}{84}-\frac{1}{82}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com