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Question Number 45932 by Tawa1 last updated on 18/Oct/18

$$\mathrm{Show}\mathrm{that}:\frac{1.2^2+2.3^2+...+\mathrm{n}{(\mathrm{n}+1)}^2}{1^2.2+2^2.3+...+{\mathrm{n}}^2(\mathrm{n}+1)}=\frac{3\mathrm{n}+5}{3\mathrm{n}+1}$$

Commented by math khazana by abdo last updated on 19/Oct/18

$$S_n=\frac{\sum _{k=1}^{n}k{(k+1)}^2}{\sum _{k=1}^n{k}^2(k+1)}=\frac{\sum _{k=1}^{n}k(k^2+2k+1)}{\sum _{k=1}^n{k}^3+\sum _k^n{k}^2}$$$$$$$$=\frac{\sum _{k=1}^n{k}^3+2\sum _{k=1}^n{k}^2+\sum _{k=1}^{n}k}{\sum _{k=1}^n{k}^3+\sum _{k=1}^n{k}^2}$$$$=\frac{\frac{n^2{(n+1)}^2}{4}+2\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{\frac{n^2{(n+1)}^2}{4}+\frac{n(n+1)(2n+1)}{6}}$$$$=\frac{\frac{n(n+1)}{2}+\frac{2}{3}(2n+1)+1}{\frac{n(n+1)}{2}+\frac{2n+1}{3}}$$$$=\frac{3n(n+1)+4(2n+1)+6}{3n(n+1)+4n+2}$$$$=\frac{3n^2+3n+8n+4+6}{3n^2+3n+4n+2}=\frac{3n^2+11n+10}{3n^2+7n+2}$$$$rootsof3n^2+7n+2$$$$\mathrm{\Delta }=49-4.3.2=49-24=25\Rightarrow n_1=\frac{-7+5}{6}=-\frac{1}{3}$$$$n_2=\frac{-7-5}{6}=-2$$$$rootsof3n^2+11n+10\to \mathrm{\Delta }=121-4.3.10$$$$=121-120=1\Rightarrow n_3=\frac{-11+1}{6}=-\frac{5}{3}$$$$n_3=\frac{-11-1}{6}=-2\Rightarrow$$$$S_n=\frac{3(n+\frac{1}{3})(n+2)}{3(n+\frac{5}{3})(n+2)}=\frac{3n+1}{3n+5}andthereisaerror$$$$attheQ.$$

Commented by math khazana by abdo last updated on 19/Oct/18

$$erroratthefinalline{S}_n=\frac{3(n+\frac{5}{3})(n+2)}{3(n+\frac{1}{3})(n+2)}$$$$=\frac{3n+5}{3n+1}andtherisnoerrorattheQ...$$

Commented by Tawa1 last updated on 19/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 19/Oct/18

$$youarewelcomesir$$

Answered by math1967 last updated on 19/Oct/18

$$S_n=1.2^2+2.3^2+.....+n{(n+1)}^2$$$$=t_1+t_2.......+n^3+2n^2+n$$$$\therefore t_1+t_2+......t_n=(1^3+2^3+...n^3)+2(1^2+2^2+..+n^2)+$$$$(1+2+..n)$$$$\therefore S_n=\frac{n^2{(n+1)}^2}{4}+2\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$$$$=\frac{n(n+1)(3n+5)(n+2)}{12}$$$$Again{1}^2.2+2^2.3+...n^2(n+1)$$$$\therefore t_n=n^3+n^2$$$$\therefore S_n=(1^3+2^3+....+n^3)+(1^2+2^2+..+n^2)$$$$=\frac{n^2{(n+1)}^2}{4}+\frac{n(n+1)(2n+1)}{6}$$$$=\frac{n(n+1)(n+2)(3n+1)}{12}$$$$\therefore \frac{1.2^2+2.3^2+....n{(n+1)}^2}{1^2.2+2^2.3+.....n^2(n+1)}$$$$\frac{n(n+1)(3n+5)(n+2)}{12}\times \frac{12}{n(n+1)(n+2)(3n+1)}$$$$=\frac{3n+5}{3n+1}\left(proved\right)$$

Commented by Tawa1 last updated on 19/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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