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Question Number 8558 by Sopheak last updated on 16/Oct/16

Commented by FilupSmith last updated on 16/Oct/16

$$S_n=\underset{k=1}{\sum ^n}\frac{k^4}{(2k-1)(2k+1)}$$$$\frac{1}{(2k-1)(2k+1)}=\frac{A}{2k-1}+\frac{B}{2k+1}$$$$1=A(2k+1)+B(2k-1)$$$$2kA+A+2kB-B=1$$$$2k(A+B)+A-B=1$$$$A+B=0$$$$A-B=1$$$$A=-B$$$$-2B=1\Rightarrow B=-\frac{1}{2}$$$$\therefore A=\frac{1}{2}$$$$$$$$\therefore \frac{1}{(2k-1)(2k+1)}=\frac{1}{2(2k-1)}+\frac{1}{2(2k+1)}$$$$\therefore \underset{k=1}{\sum ^n}\frac{k^4}{(2k-1)(2k+1)}=\frac{1}{2}\underset{k=1}{\sum ^n}{k}^4(\frac{1}{(2k-1)}+\frac{1}{(2k+1)})$$$$2S_n=\underset{k=1}{\sum ^n}\frac{k^4}{2k-1}+\underset{k=1}{\sum ^n}\frac{k^4}{2k+1}$$$$continue$$

Commented by FilupSmith last updated on 16/Oct/16

$$\mathrm{For}\mathrm{reference},\mathrm{according}\mathrm{to}\mathrm{wolframAlpha}:$$$$S_n=\frac{n^4+2n^3+2n^2+n}{12n+6}$$

Commented by prakash jain last updated on 16/Oct/16

$$\frac{k^4}{(2k-1)(2k+1)}=\frac{1}{16}\left[\frac{16k^4-1+1}{(2k-1)(2k+1)}\right]$$$$=\frac{1}{16}[\frac{(4k^2-1)(4k^2+1)}{(2k-1)(2k+1)}+\frac{1}{(2k-1)(2k+1)}]$$$$=\frac{1}{16}[4k^2+1+\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})]$$$$\underset{k=1}{\sum ^n}\frac{k^4}{(2k-1)(2k+1)}=\frac{1}{16}\underset{k=1}{\sum ^n}4k^2+\frac{1}{16}\underset{k=1}{\sum ^n}1$$$$+\underset{k=1}{\sum ^n}(\frac{1}{2k-1}-\frac{1}{2k+1})$$$$\underset{k=1}{\sum ^n}(\frac{1}{2k-1}-\frac{1}{2k+1})=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}..+\frac{1}{2n-1}-\frac{1}{2n+1}$$$$=1-\frac{1}{2n+1}=\frac{2n}{2n+1}$$$$\frac{1}{32}\underset{k=1}{\sum ^n}(\frac{1}{2k-1}-\frac{1}{2k+1})=\frac{n}{16(2n+1)}$$$$\frac{1}{16}\underset{k=1}{\sum ^n}4k^2=\frac{n(n+1)(2n+1)}{24}$$$$\frac{1}{16}\underset{k=1}{\sum ^n}1=\frac{n}{16}$$$$\underset{k=1}{\sum ^n}\frac{k^4}{(2k-1)(2k+1)}=\frac{n(n+1)(2n+1)}{24}+\frac{n}{16}+\frac{1}{16}\left(\frac{n}{2n+1}\right)$$$$=\frac{2n(n+1){(2n+1)}^2+3n(2n+1)+3n}{48(2n+1)}$$$$$$$$=\frac{(2n^2+2n)(4n^2+4n+1)+(6n^2+3n)+3n}{48(2n+1)}$$$$=\frac{8n^4+8n^3+8n^3+8n^2+2n^2+2n+6n^2+3n+3n}{48(2n+1)}$$$$=\frac{8n^4+16n^3+16n^2+8n}{48(2n+1)}=\frac{n^4+2n^3+2n^2+n}{6(2n+1)}$$

Answered by prakash jain last updated on 16/Oct/16

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