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Question Number 75818 by naka3546 last updated on 18/Dec/19

Given  the  increasing  sequence :  1, 4, 8, 13, ...  a. Find  U_(2019)   b. Find  S_(2019)   U_n   is  nth−term  of  the  sequence  S_n   is  sum  of  n − term  of  the  sequence  Arithmetic  Sequence  Degree  Two

$${Given}\:\:{the}\:\:{increasing}\:\:{sequence}\:: \\ $$$$\mathrm{1},\:\mathrm{4},\:\mathrm{8},\:\mathrm{13},\:... \\ $$$${a}.\:{Find}\:\:{U}_{\mathrm{2019}} \\ $$$${b}.\:{Find}\:\:{S}_{\mathrm{2019}} \\ $$$${U}_{{n}} \:\:{is}\:\:{nth}−{term}\:\:{of}\:\:{the}\:\:{sequence} \\ $$$${S}_{{n}} \:\:{is}\:\:{sum}\:\:{of}\:\:{n}\:−\:{term}\:\:{of}\:\:{the}\:\:{sequence} \\ $$$${Arithmetic}\:\:{Sequence}\:\:{Degree}\:\:{Two} \\ $$

Commented by MJS last updated on 18/Dec/19

the sequence is not unique  1+3=4  4+4=8  8+5=13  the next terms are  13+6=19  19+7=26  ...    1+4+8=13  the next terms are  4+8+13=25  8+13+25=46  ...    it′s possible to find a zillion more...

$$\mathrm{the}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{not}\:\mathrm{unique} \\ $$$$\mathrm{1}+\mathrm{3}=\mathrm{4} \\ $$$$\mathrm{4}+\mathrm{4}=\mathrm{8} \\ $$$$\mathrm{8}+\mathrm{5}=\mathrm{13} \\ $$$$\mathrm{the}\:\mathrm{next}\:\mathrm{terms}\:\mathrm{are} \\ $$$$\mathrm{13}+\mathrm{6}=\mathrm{19} \\ $$$$\mathrm{19}+\mathrm{7}=\mathrm{26} \\ $$$$... \\ $$$$ \\ $$$$\mathrm{1}+\mathrm{4}+\mathrm{8}=\mathrm{13} \\ $$$$\mathrm{the}\:\mathrm{next}\:\mathrm{terms}\:\mathrm{are} \\ $$$$\mathrm{4}+\mathrm{8}+\mathrm{13}=\mathrm{25} \\ $$$$\mathrm{8}+\mathrm{13}+\mathrm{25}=\mathrm{46} \\ $$$$... \\ $$$$ \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{zillion}\:\mathrm{more}... \\ $$

Commented by MJS last updated on 18/Dec/19

not really  in some cases if we do not know what we are  supposed to do we cannot give a proper  answer

$$\mathrm{not}\:\mathrm{really} \\ $$$$\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{if}\:\mathrm{we}\:\mathrm{do}\:\mathrm{not}\:\mathrm{know}\:\mathrm{what}\:\mathrm{we}\:\mathrm{are} \\ $$$$\mathrm{supposed}\:\mathrm{to}\:\mathrm{do}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{give}\:\mathrm{a}\:\mathrm{proper} \\ $$$$\mathrm{answer} \\ $$

Commented by $@ty@m123 last updated on 18/Dec/19

Its exaggeration.

$${Its}\:{exaggeration}. \\ $$

Answered by $@ty@m123 last updated on 18/Dec/19

S_n =1+4+8+13+.....+U_n   S_n =       1+4+8+13+.....+U_n   Subtracting,  0=1+3+4+5+6+.....+(U_n −U_(n−1) )−U_n   U_n =1+3+4+5+6+.....+(U_n −U_(n−1) )  U_n =1+2+3+4+5+6+.....+(U_n −U_(n−1) )−2  U_n =(((n−1).n)/2) −2 ......(1)  U_(2019) =((2018×2019)/2)−2   U_(2019) =1009×2019−2=2037169  S_n =ΣU_n   =(1/2)(Σn^2 −Σn)−Σ2  =(1/2){((n(n+1)(2n+1))/6)−((n(n+1))/2)}−2n  =((n(n+1))/4){(((2n+1))/3)−1}−2n  Put n=2019 & calculate.

$${S}_{{n}} =\mathrm{1}+\mathrm{4}+\mathrm{8}+\mathrm{13}+.....+{U}_{{n}} \\ $$$${S}_{{n}} =\:\:\:\:\:\:\:\mathrm{1}+\mathrm{4}+\mathrm{8}+\mathrm{13}+.....+{U}_{{n}} \\ $$$${Subtracting}, \\ $$$$\mathrm{0}=\mathrm{1}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+.....+\left({U}_{{n}} −{U}_{{n}−\mathrm{1}} \right)−{U}_{{n}} \\ $$$${U}_{{n}} =\mathrm{1}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+.....+\left({U}_{{n}} −{U}_{{n}−\mathrm{1}} \right) \\ $$$${U}_{{n}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+.....+\left({U}_{{n}} −{U}_{{n}−\mathrm{1}} \right)−\mathrm{2} \\ $$$${U}_{{n}} =\frac{\left({n}−\mathrm{1}\right).{n}}{\mathrm{2}}\:−\mathrm{2}\:......\left(\mathrm{1}\right) \\ $$$${U}_{\mathrm{2019}} =\frac{\mathrm{2018}×\mathrm{2019}}{\mathrm{2}}−\mathrm{2}\: \\ $$$${U}_{\mathrm{2019}} =\mathrm{1009}×\mathrm{2019}−\mathrm{2}=\mathrm{2037169} \\ $$$${S}_{{n}} =\Sigma{U}_{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\Sigma{n}^{\mathrm{2}} −\Sigma{n}\right)−\Sigma\mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}−\mathrm{2}{n} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}}\left\{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{3}}−\mathrm{1}\right\}−\mathrm{2}{n} \\ $$$${Put}\:{n}=\mathrm{2019}\:\&\:{calculate}. \\ $$

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