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Question Number 5125 by Rojaye Shegz last updated on 16/Apr/16

Commented by Rojaye Shegz last updated on 16/Apr/16

Sorry, that was an error. (p/q) is the {(1/2)(p+q−1)(p+q−2)+p}th term

$$\mathrm{Sorry},\:\mathrm{that}\:\mathrm{was}\:\mathrm{an}\:\mathrm{error}. \\ $$$$\frac{\mathrm{p}}{\mathrm{q}}\:\mathrm{is}\:\mathrm{the}\:\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{p}+\mathrm{q}−\mathrm{1}\right)\left(\mathrm{p}+\mathrm{q}−\mathrm{2}\right)+\mathrm{p}\right\}\mathrm{th}\:\mathrm{term} \\ $$

Commented by 123456 last updated on 17/Apr/16

(1/1)⇒1+1=2(1) (1/2),(2/1)⇒1+2=3(2) (1/3),(2/2),(3/1)⇒1+3=4(3) (1/4),(2/3),(3/2),(4/1)⇒1+4=5(4) ⋮ (1/q),(2/(q−1)),(3/(q−2)),…,((q−2)/3),((q−1)/2),(q/1)⇒p+q=k(p+q−1) Σ_(i=1) ^(k−1) (i−1)=(((k−1−1+1)(k−1−1+1−1))/2) =(((k−1)(k−2))/2) =(((a+b−1)(a+b−2))/2) (1/q),(2/(q−1)),...,(p/q),...,((q−1)/2),(q/1)⇒p (((a+b−1)(a+b−2))/2)+p

$$\frac{\mathrm{1}}{\mathrm{1}}\Rightarrow\mathrm{1}+\mathrm{1}=\mathrm{2}\left(\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{1}}\Rightarrow\mathrm{1}+\mathrm{2}=\mathrm{3}\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{1}}\Rightarrow\mathrm{1}+\mathrm{3}=\mathrm{4}\left(\mathrm{3}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{4}}{\mathrm{1}}\Rightarrow\mathrm{1}+\mathrm{4}=\mathrm{5}\left(\mathrm{4}\right) \\ $$$$\vdots \\ $$$$\frac{\mathrm{1}}{{q}},\frac{\mathrm{2}}{{q}−\mathrm{1}},\frac{\mathrm{3}}{{q}−\mathrm{2}},\ldots,\frac{{q}−\mathrm{2}}{\mathrm{3}},\frac{{q}−\mathrm{1}}{\mathrm{2}},\frac{{q}}{\mathrm{1}}\Rightarrow{p}+{q}={k}\left({p}+{q}−\mathrm{1}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}\left({i}−\mathrm{1}\right)=\frac{\left({k}−\mathrm{1}−\mathrm{1}+\mathrm{1}\right)\left({k}−\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{\left({k}−\mathrm{1}\right)\left({k}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$=\frac{\left({a}+{b}−\mathrm{1}\right)\left({a}+{b}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{q}},\frac{\mathrm{2}}{{q}−\mathrm{1}},...,\frac{{p}}{{q}},...,\frac{{q}−\mathrm{1}}{\mathrm{2}},\frac{{q}}{\mathrm{1}}\Rightarrow{p} \\ $$$$\frac{\left({a}+{b}−\mathrm{1}\right)\left({a}+{b}−\mathrm{2}\right)}{\mathrm{2}}+{p} \\ $$

Commented by Rojaye Shegz last updated on 17/Apr/16

I dont understand your solution but im gaining insight on how it should be solved. Please explain Thanks^+

$$\mathrm{I}\:\mathrm{dont}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{solution} \\ $$$$\mathrm{but}\:\mathrm{im}\:\mathrm{gaining}\:\mathrm{insight}\:\mathrm{on}\:\mathrm{how}\:\mathrm{it}\:\mathrm{should} \\ $$$$\mathrm{be}\:\mathrm{solved}.\:\mathrm{Please}\:\mathrm{explain} \\ $$$$\mathrm{Thanks}^{+} \\ $$

Answered by Yozzii last updated on 16/Apr/16

For u_3 =(2/1)=(p/q) ⇒q=1 & p=2. ∴ 0.5(3−1)(3−2)+1=0.5×2+1=1+1=2≠3 For u_2 =(1/2)⇒p=1,q=2 ∴ 0.5(3−1)(3−2)+2=3≠2.

$${For}\:{u}_{\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{1}}=\frac{{p}}{{q}} \\ $$$$\Rightarrow{q}=\mathrm{1}\:\&\:{p}=\mathrm{2}. \\ $$$$\therefore\:\mathrm{0}.\mathrm{5}\left(\mathrm{3}−\mathrm{1}\right)\left(\mathrm{3}−\mathrm{2}\right)+\mathrm{1}=\mathrm{0}.\mathrm{5}×\mathrm{2}+\mathrm{1}=\mathrm{1}+\mathrm{1}=\mathrm{2}\neq\mathrm{3} \\ $$$${For}\:{u}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{p}=\mathrm{1},{q}=\mathrm{2} \\ $$$$\therefore\:\mathrm{0}.\mathrm{5}\left(\mathrm{3}−\mathrm{1}\right)\left(\mathrm{3}−\mathrm{2}\right)+\mathrm{2}=\mathrm{3}\neq\mathrm{2}. \\ $$

Answered by 123456 last updated on 17/Apr/16

lets organize the sequence on this way (1/1) (1/2),(2/1) (1/3),(2/2),(3/1) (1/4),(2/3),(3/2),(4/1) ⋮ each colum have a+b=comstant with (a/b), and its is i+1(number of colum plus one) lets have (p/q), conting all elements before this colum the colum of p/q is p+q−1, so one before is p+q−2 Σ_(i=1) ^(p+q−2) i=(((p+q−2)(p+q−2+1))/2)(P.A) =(((p+q−2)(p+q−1))/2) now lets count element at colum p+q−1 (1/q),(2/(q−1)),...,(p/q) as you can see, at p/q we have p elements so the total of elements is (including it) (((p+q−2)(p+q−1))/2)+p wich make it (((p+q−2)(p+q−1))/2)+p nth element

$$\mathrm{lets}\:\mathrm{organize}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{on}\:\mathrm{this}\:\mathrm{way} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{4}}{\mathrm{1}} \\ $$$$\vdots \\ $$$$\mathrm{each}\:\mathrm{colum}\:\mathrm{have}\:{a}+{b}=\mathrm{comstant}\:\mathrm{with} \\ $$$$\frac{{a}}{{b}},\:\mathrm{and}\:\mathrm{its}\:\mathrm{is}\:{i}+\mathrm{1}\left(\mathrm{number}\:\mathrm{of}\:\mathrm{colum}\:\mathrm{plus}\:\mathrm{one}\right) \\ $$$$\mathrm{lets}\:\mathrm{have}\:\frac{{p}}{{q}},\:\mathrm{conting}\:\mathrm{all}\:\mathrm{elements}\:\mathrm{before} \\ $$$$\mathrm{this}\:\mathrm{colum} \\ $$$$\mathrm{the}\:\mathrm{colum}\:\mathrm{of}\:{p}/{q}\:\mathrm{is}\:{p}+{q}−\mathrm{1},\:\mathrm{so}\:\mathrm{one}\:\mathrm{before}\:\mathrm{is} \\ $$$${p}+{q}−\mathrm{2} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{p}+{q}−\mathrm{2}} {\sum}}{i}=\frac{\left({p}+{q}−\mathrm{2}\right)\left({p}+{q}−\mathrm{2}+\mathrm{1}\right)}{\mathrm{2}}\left({P}.{A}\right) \\ $$$$=\frac{\left({p}+{q}−\mathrm{2}\right)\left({p}+{q}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{lets}\:\mathrm{count}\:\mathrm{element}\:\mathrm{at}\:\mathrm{colum}\:{p}+{q}−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{q}},\frac{\mathrm{2}}{{q}−\mathrm{1}},...,\frac{{p}}{{q}} \\ $$$$\mathrm{as}\:\mathrm{you}\:\mathrm{can}\:\mathrm{see},\:\mathrm{at}\:{p}/{q}\:{we}\:{have}\:{p}\:\mathrm{elements} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{total}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{is}\:\left(\mathrm{including}\:\mathrm{it}\right) \\ $$$$\frac{\left({p}+{q}−\mathrm{2}\right)\left({p}+{q}−\mathrm{1}\right)}{\mathrm{2}}+{p} \\ $$$$\mathrm{wich}\:\mathrm{make}\:\mathrm{it} \\ $$$$\frac{\left({p}+{q}−\mathrm{2}\right)\left({p}+{q}−\mathrm{1}\right)}{\mathrm{2}}+{p}\:\:\mathrm{nth}\:\mathrm{element} \\ $$

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