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Question Number 206367 by hardmath last updated on 12/Apr/24

$$\mathrm{Find}:$$$$\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+...\frac{1}{720}=?$$

Commented by TheHoneyCat last updated on 12/Apr/24

Can you please explain what the "..." are?

Commented by TonyCWX08 last updated on 13/Apr/24

$$Canyougiveuntilthefourthterm???$$$$Oratleastgivemethenumberofterms???$$

Answered by mr W last updated on 13/Apr/24

$$T_k=\frac{1}{k(k+1)(k+2)}withk=1,2,3,...,8$$$$tosumup8termsyoucanuseyour$$$$calculator:$$$$\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\frac{1}{210}+\frac{1}{336}+\frac{1}{504}+\frac{1}{720}=\frac{11}{45}$$$$$$$${it}^{\prime }smorechallengingtofind$$$$\underset{k=1}{\sum ^n}\frac{1}{k(k+1)(k+2)}=?$$

Commented by mr W last updated on 13/Apr/24

$$solution:$$$$\frac{1}{k(k+1)(k+2)}=\frac{A}{k(k+1)}+\frac{B}{(k+1)(k+2)}$$$$1=A(k+2)+Bk$$$$2A=1\Rightarrow A=\frac{1}{2}$$$$A+B=0\Rightarrow B=-\frac{1}{2}$$$$\underset{k=1}{\sum ^n}\frac{1}{k(k+1)(k+2)}$$$$=\frac{1}{2}\underset{k=1}{\sum ^n}(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)})$$$$=\frac{1}{2}(\frac{1}{1\times 2}-\frac{1}{(n+1)(n+2)})$$$$=\frac{n(n+3)}{4(n+1)(n+2)}$$$$withn=8:\frac{11}{45}$$

Commented by hardmath last updated on 13/Apr/24

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

Answered by MATHEMATICSAM last updated on 13/Apr/24

$$\mathrm{We}\mathrm{can}\mathrm{see}\mathrm{each}\mathrm{term}\mathrm{in}\mathrm{terms}\mathrm{of}n\mathrm{is}$$$$\frac{1}{n^3-n}=\frac{1}{2}\times \frac{2}{n(n+1)(n-1)}$$$$=\frac{1}{2}\times \frac{(n+1)-(n-1)}{n(n+1)(n-1)}$$$$=\frac{1}{2}[\frac{1}{(n-1)n}-\frac{1}{n(n+1)}]$$$$\mathrm{where}n=2,3,4,...,9\mathrm{respectively}.$$$$\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+....+\frac{1}{720}$$$$=\frac{1}{2}[\frac{1}{1\times 2}-\cancel{\frac{1}{2\times 3}}+\cancel{\frac{1}{2\times 3}}-\cancel{\frac{1}{3\times 4}}+\cancel{\frac{1}{3\times 4}}-\cancel{\frac{1}{4\times 5}}+.....+\cancel{\frac{1}{8\times 9}}-\frac{1}{9\times 10}]$$$$=\frac{1}{2}[\frac{1}{2}-\frac{1}{90}]$$$$=\frac{1}{2}\times \frac{44}{90}=\frac{11}{45}\left(\mathrm{Ans}\right)$$

Answered by BaliramKumar last updated on 13/Apr/24

$$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+......+\frac{1}{8\times 9\times 10}=?$$$$\mathrm{d}=3-1=4-2=5-3=2$$$$\frac{1}{\mathrm{d}}[\frac{1}{1\times 2}-\frac{1}{9\times 10}]$$$$\frac{1}{2}[\frac{1}{1\times 2}-\frac{1}{9\times 10}]=\frac{1}{2}\left[\frac{44}{90}\right]=\frac{11}{45}$$

Answered by Frix last updated on 13/Apr/24

$$\mathrm{I}\mathrm{know}...$$$$\mathrm{but}\mathrm{this}\mathrm{is}\mathrm{also}\mathrm{a}\mathrm{valid}\mathrm{answer}:$$$$f\left(x\right)=y=c_3{x}^3+c_2{x}^2+c_{1}x+d$$$$\mathrm{We}\mathrm{have}$$$$x=1\wedge y=\frac{1}{6}$$$$x=2\wedge y=\frac{1}{24}$$$$x=3\wedge y=\frac{1}{60}$$$$\mathrm{and}\mathrm{since}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{number}\mathrm{of}$$$$\mathrm{elements}\mathrm{between}\frac{1}{60}\mathrm{and}\frac{1}{720}$$$$x=n\wedge y=\frac{1}{720}$$$$\mathrm{which}\mathrm{leads}\mathrm{to}$$$$f\left(x\right)=-\frac{36n^2-198n+281}{720(n-3)(n-2)(n-1)}{x}^3+$$$$+\frac{6n^3-132n+245}{120(n-3)(n-2)(n-1)}{x}^2-$$$$-\frac{11(18n^3-72n^2+173)}{720(n-3)(n-2)(n-1)}x+$$$$+\frac{47n^3-246n^2+319n-1}{120(n-3)(n-2)(n-1)}$$$$\Rightarrow$$$$\underset{x=1}{\sum ^n}f\left(x\right)=\frac{n(12n^2-126n+475)}{2880}$$$$$$$$$$$$\mathrm{If}\mathrm{we}\mathrm{use}\frac{1}{f\left(x\right)}={ax}^3+{bx}^2+cx+d\mathrm{with}\mathrm{the}$$$$\mathrm{same}\mathrm{data}\mathrm{we}\mathrm{get}$$$$\frac{1}{f\left(x\right)}=-\frac{3(3n^2-3n-238)}{(n-3)(n-2)(n-1)}{x}^3+$$$$+\frac{9(n^3+5n-482)}{(n-3)(n-2)(n-1)}{x}^2-$$$$-\frac{3(3n^3+15n^2-2636)}{(n-3)(n-2)(n-1)}x+$$$$+\frac{6(n-8)(n^2+11n+90)}{(n-3)(n-2)(n-1)}$$$$\mathrm{and}\mathrm{the}\mathrm{sum}\underset{x=1}{\sum ^n}f\left(x\right)\mathrm{cannot}\mathrm{be}\mathrm{expressed}$$$$\mathrm{in}\mathrm{a}\mathrm{closed}\mathrm{form}.$$$$\mathrm{Anyway}\mathrm{for}n=8\mathrm{we}\mathrm{get}\mathrm{tbe}\mathrm{nice}f\left(x\right)\mathrm{my}$$$$\mathrm{fellows}\mathrm{here}\mathrm{stumbled}\mathrm{upon}.$$

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