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Question Number 206367 by hardmath last updated on 12/Apr/24
Find: (1/6) + (1/(24)) + (1/(60)) + ... + ... (1/(720)) = ?
$$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{24}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{60}}\:\:+\:…\:+\:…\:\:\frac{\mathrm{1}}{\mathrm{720}}\:=\:? \\ $$
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
Can you give until the fourth term??? Or at least give me the number of terms???
$${Can}\:{you}\:{give}\:{until}\:{the}\:{fourth}\:{term}??? \\ $$$${Or}\:{at}\:{least}\:{give}\:{me}\:{the}\:{number}\:{of}\:{terms}??? \\ $$
Answered by mr W last updated on 13/Apr/24
T_k =(1/(k(k+1)(k+2))) with k=1, 2, 3, ..., 8 to sum up 8 terms you can use your calculator: (1/6)+(1/(24))+(1/(60))+(1/(120))+(1/(210))+(1/(336))+(1/(504))+(1/(720))=((11)/(45)) it′s more challenging to find Σ_(k=1) ^n (1/(k(k+1)(k+2)))=?
$${T}_{{k}} =\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\:{with}\:{k}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…,\:\mathrm{8} \\ $$$${to}\:{sum}\:{up}\:\mathrm{8}\:{terms}\:{you}\:{can}\:{use}\:{your} \\ $$$${calculator}: \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{60}}+\frac{\mathrm{1}}{\mathrm{120}}+\frac{\mathrm{1}}{\mathrm{210}}+\frac{\mathrm{1}}{\mathrm{336}}+\frac{\mathrm{1}}{\mathrm{504}}+\frac{\mathrm{1}}{\mathrm{720}}=\frac{\mathrm{11}}{\mathrm{45}} \\ $$$$ \\ $$$${it}'{s}\:{more}\:{challenging}\:{to}\:{find} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=? \\ $$
Commented by mr W last updated on 13/Apr/24
solution: (1/(k(k+1)(k+2)))=(A/(k(k+1)))+(B/((k+1)(k+2))) 1=A(k+2)+Bk 2A=1 ⇒A=(1/2) A+B=0 ⇒B=−(1/2) Σ_(k=1) ^n (1/(k(k+1)(k+2))) =(1/2)Σ_(k=1) ^n ((1/(k(k+1)))−(1/((k+1)(k+2)))) =(1/2)((1/(1×2))−(1/((n+1)(n+2)))) =((n(n+3))/(4(n+1)(n+2))) with n=8: ((11)/(45))
$${solution}: \\ $$$$\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\frac{{A}}{{k}\left({k}+\mathrm{1}\right)}+\frac{{B}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$\mathrm{1}={A}\left({k}+\mathrm{2}\right)+{Bk} \\ $$$$\mathrm{2}{A}=\mathrm{1}\:\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}+{B}=\mathrm{0}\:\Rightarrow{B}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{4}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$${with}\:{n}=\mathrm{8}:\:\:\:\frac{\mathrm{11}}{\mathrm{45}} \\ $$
Commented by hardmath last updated on 13/Apr/24
cool dear professor thank you
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by MATHEMATICSAM last updated on 13/Apr/24
We can see each term in terms of n is (1/(n^3 − n)) = (1/2) × (2/(n(n + 1)(n − 1))) = (1/2) × (((n + 1)−(n − 1))/(n(n + 1)(n − 1))) = (1/2) [(1/((n − 1)n)) − (1/(n(n + 1)))] where n = 2, 3, 4,..., 9 respectively. (1/6) + (1/(24)) + (1/(60)) + .... + (1/(720)) = (1/2) [(1/(1 × 2)) − (1/(2 × 3)) + (1/(2 × 3)) − (1/(3 × 4)) + (1/(3 × 4)) − (1/(4 × 5)) + ..... + (1/(8 × 9)) − (1/(9 × 10))] = (1/2)[(1/2) − (1/(90))] = (1/2) × ((44)/(90)) = ((11)/(45)) (Ans)
$$\mathrm{We}\:\mathrm{can}\:\mathrm{see}\:\mathrm{each}\:\mathrm{term}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}\:\mathrm{is} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{3}} \:−\:{n}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{\mathrm{2}}{{n}\left({n}\:+\:\mathrm{1}\right)\left({n}\:−\:\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{\left({n}\:+\:\mathrm{1}\right)−\left({n}\:−\:\mathrm{1}\right)}{{n}\left({n}\:+\:\mathrm{1}\right)\left({n}\:−\:\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\left({n}\:−\:\mathrm{1}\right){n}}\:−\:\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)}\right] \\ $$$$\mathrm{where}\:{n}\:=\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},…,\:\mathrm{9}\:\mathrm{respectively}. \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{24}}\:+\:\frac{\mathrm{1}}{\mathrm{60}}\:+\:….\:+\:\frac{\mathrm{1}}{\mathrm{720}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\mathrm{1}\:×\:\mathrm{2}}\:−\:\cancel{\frac{\mathrm{1}}{\mathrm{2}\:×\:\mathrm{3}}}\:+\:\cancel{\frac{\mathrm{1}}{\mathrm{2}\:×\:\mathrm{3}}}\:−\:\cancel{\frac{\mathrm{1}}{\mathrm{3}\:×\:\mathrm{4}}}\:+\:\cancel{\frac{\mathrm{1}}{\mathrm{3}\:×\:\mathrm{4}}}\:−\:\cancel{\frac{\mathrm{1}}{\mathrm{4}\:×\:\mathrm{5}}}\:+\:…..\:+\:\cancel{\frac{\mathrm{1}}{\mathrm{8}\:×\:\mathrm{9}}}\:−\:\frac{\mathrm{1}}{\mathrm{9}\:×\:\mathrm{10}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{90}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{\mathrm{44}}{\mathrm{90}}\:=\:\frac{\mathrm{11}}{\mathrm{45}}\:\left(\mathrm{Ans}\right) \\ $$
Answered by BaliramKumar last updated on 13/Apr/24
(1/(1×2×3)) + (1/(2×3×4)) + (1/(3×4×5)) + ... ... + (1/(8×9×10)) = ? d = 3−1=4−2= 5−3 = 2 (1/d)[(1/(1×2)) − (1/(9×10))] (1/2)[(1/(1×2)) − (1/(9×10))] = (1/2)[((44)/(90))] = ((11)/(45))
$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:\:+\:…\:…\:+\:\frac{\mathrm{1}}{\mathrm{8}×\mathrm{9}×\mathrm{10}}\:=\:?\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{d}\:=\:\mathrm{3}−\mathrm{1}=\mathrm{4}−\mathrm{2}=\:\mathrm{5}−\mathrm{3}\:=\:\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{d}}\left[\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{9}×\mathrm{10}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{9}×\mathrm{10}}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{44}}{\mathrm{90}}\right]\:=\:\frac{\mathrm{11}}{\mathrm{45}} \\ $$
Answered by Frix last updated on 13/Apr/24
I know... but this is also a valid answer: f(x)=y=c_3 x^3 +c_2 x^2 +c_1 x+d We have x=1∧y=(1/6) x=2∧y=(1/(24)) x=3∧y=(1/(60)) and since we don′t know the number of elements between (1/(60)) and (1/(720)) x=n∧y=(1/(720)) which leads to f(x)=−((36n^2 −198n+281)/(720(n−3)(n−2)(n−1)))x^3 + +((6n^3 −132n+245)/(120(n−3)(n−2)(n−1)))x^2 − −((11(18n^3 −72n^2 +173))/(720(n−3)(n−2)(n−1)))x+ +((47n^3 −246n^2 +319n−1)/(120(n−3)(n−2)(n−1))) ⇒ Σ_(x=1) ^n f(x) =((n(12n^2 −126n+475))/(2880)) If we use (1/(f(x)))=ax^3 +bx^2 +cx+d with the same data we get (1/(f(x)))=−((3(3n^2 −3n−238))/((n−3)(n−2)(n−1)))x^3 + +((9(n^3 +5n−482))/((n−3)(n−2)(n−1)))x^2 − −((3(3n^3 +15n^2 −2636))/((n−3)(n−2)(n−1)))x+ +((6(n−8)(n^2 +11n+90))/((n−3)(n−2)(n−1))) and the sum Σ_(x=1) ^n f(x) cannot be expressed in a closed form. Anyway for n=8 we get tbe nice f(x) my fellows here stumbled upon.
$$\mathrm{I}\:\mathrm{know}… \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{answer}: \\ $$$${f}\left({x}\right)={y}={c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{d} \\ $$$$\mathrm{We}\:\mathrm{have} \\ $$$${x}=\mathrm{1}\wedge{y}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${x}=\mathrm{2}\wedge{y}=\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${x}=\mathrm{3}\wedge{y}=\frac{\mathrm{1}}{\mathrm{60}} \\ $$$$\mathrm{and}\:\mathrm{since}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{elements}\:\mathrm{between}\:\frac{\mathrm{1}}{\mathrm{60}}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{720}} \\ $$$${x}={n}\wedge{y}=\frac{\mathrm{1}}{\mathrm{720}} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{36}{n}^{\mathrm{2}} −\mathrm{198}{n}+\mathrm{281}}{\mathrm{720}\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{x}^{\mathrm{3}} + \\ $$$$\:\:\:\:\:+\frac{\mathrm{6}{n}^{\mathrm{3}} −\mathrm{132}{n}+\mathrm{245}}{\mathrm{120}\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{x}^{\mathrm{2}} − \\ $$$$\:\:\:\:\:−\frac{\mathrm{11}\left(\mathrm{18}{n}^{\mathrm{3}} −\mathrm{72}{n}^{\mathrm{2}} +\mathrm{173}\right)}{\mathrm{720}\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{x}+ \\ $$$$\:\:\:\:\:+\frac{\mathrm{47}{n}^{\mathrm{3}} −\mathrm{246}{n}^{\mathrm{2}} +\mathrm{319}{n}−\mathrm{1}}{\mathrm{120}\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)} \\ $$$$\Rightarrow \\ $$$$\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}\:{f}\left({x}\right)\:=\frac{{n}\left(\mathrm{12}{n}^{\mathrm{2}} −\mathrm{126}{n}+\mathrm{475}\right)}{\mathrm{2880}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{use}\:\frac{\mathrm{1}}{{f}\left({x}\right)}={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{data}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}}{{f}\left({x}\right)}=−\frac{\mathrm{3}\left(\mathrm{3}{n}^{\mathrm{2}} −\mathrm{3}{n}−\mathrm{238}\right)}{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{x}^{\mathrm{3}} + \\ $$$$\:\:\:\:\:+\frac{\mathrm{9}\left({n}^{\mathrm{3}} +\mathrm{5}{n}−\mathrm{482}\right)}{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{x}^{\mathrm{2}} − \\ $$$$\:\:\:\:\:−\frac{\mathrm{3}\left(\mathrm{3}{n}^{\mathrm{3}} +\mathrm{15}{n}^{\mathrm{2}} −\mathrm{2636}\right)}{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{x}+ \\ $$$$\:\:\:\:\:+\frac{\mathrm{6}\left({n}−\mathrm{8}\right)\left({n}^{\mathrm{2}} +\mathrm{11}{n}+\mathrm{90}\right)}{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}\:{f}\left({x}\right)\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{expressed} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}. \\ $$$$\mathrm{Anyway}\:\mathrm{for}\:{n}=\mathrm{8}\:\mathrm{we}\:\mathrm{get}\:\mathrm{tbe}\:\mathrm{nice}\:{f}\left({x}\right)\:\mathrm{my} \\ $$$$\mathrm{fellows}\:\mathrm{here}\:\mathrm{stumbled}\:\mathrm{upon}. \\ $$

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