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Question Number 86088 by Serlea last updated on 27/Mar/20
Find the sum of the series below: 1+2+3−4−5−6+7+8+9−10−11−12+13+14+15...−3020
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\:\mathrm{below}: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}−\mathrm{4}−\mathrm{5}−\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}−\mathrm{10}−\mathrm{11}−\mathrm{12}+\mathrm{13}+\mathrm{14}+\mathrm{15}…−\mathrm{3020} \\ $$
Commented by Kunal12588 last updated on 27/Mar/20
did yoy forgot 10? I think series should be like 1+2+3−4−5−6+7+8+9−10−11−12+13+...+3020
$${did}\:{yoy}\:{forgot}\:\mathrm{10}? \\ $$$${I}\:{think}\:{series}\:{should}\:{be}\:{like} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}−\mathrm{4}−\mathrm{5}−\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}−\mathrm{10}−\mathrm{11}−\mathrm{12}+\mathrm{13}+…+\mathrm{3020} \\ $$
Commented by Serlea last updated on 27/Mar/20
Merci
$$\mathrm{Merci} \\ $$
Commented by Prithwish Sen 1 last updated on 27/Mar/20
I think rhe question is 1+2+3−4−5−6+7+8+9−10−11−12...... −3016−3017−3018 =(1−4)+(2−5)+(3−6)+......(3013−3016) +(3014−3017)+(3015−3018) =(−3)×1509 = −4527 please check.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{rhe}\:\mathrm{question}\:\mathrm{is}\: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}−\mathrm{4}−\mathrm{5}−\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}−\mathrm{10}−\mathrm{11}−\mathrm{12}…… \\ $$$$−\mathrm{3016}−\mathrm{3017}−\mathrm{3018} \\ $$$$=\left(\mathrm{1}−\mathrm{4}\right)+\left(\mathrm{2}−\mathrm{5}\right)+\left(\mathrm{3}−\mathrm{6}\right)+……\left(\mathrm{3013}−\mathrm{3016}\right) \\ $$$$+\left(\mathrm{3014}−\mathrm{3017}\right)+\left(\mathrm{3015}−\mathrm{3018}\right) \\ $$$$=\left(−\mathrm{3}\right)×\mathrm{1509}\:=\:−\mathrm{4527} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Commented by Kunal12588 last updated on 28/Mar/20
see sir next term would be ...+(3−6)+(4−7)+(5−8)+... and these terms are not needed
$${see}\:{sir}\:{next}\:{term}\:{would}\:{be} \\ $$$$…+\left(\mathrm{3}−\mathrm{6}\right)+\left(\mathrm{4}−\mathrm{7}\right)+\left(\mathrm{5}−\mathrm{8}\right)+… \\ $$$${and}\:{these}\:{terms}\:{are}\:{not}\:{needed} \\ $$
Commented by Prithwish Sen 1 last updated on 28/Mar/20
I think you are making a mistake , actually it is (1−4)+(2−5)+(3−6)+(7−10)+(8−11)+ (9−12)+(13−16)+(14−17)+(15−18)+.. amd so on. look here nothing is repeated. You have to take six numbers as a group like 1+2+3−4−5−6 then 7+8+9−10−11−12 then 13+14+15−16−17−18 and so on.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{are}\:\mathrm{making}\:\mathrm{a}\:\mathrm{mistake}\:, \\ $$$$\mathrm{actually}\:\mathrm{it}\:\mathrm{is} \\ $$$$\left(\mathrm{1}−\mathrm{4}\right)+\left(\mathrm{2}−\mathrm{5}\right)+\left(\mathrm{3}−\mathrm{6}\right)+\left(\mathrm{7}−\mathrm{10}\right)+\left(\mathrm{8}−\mathrm{11}\right)+ \\ $$$$\left(\mathrm{9}−\mathrm{12}\right)+\left(\mathrm{13}−\mathrm{16}\right)+\left(\mathrm{14}−\mathrm{17}\right)+\left(\mathrm{15}−\mathrm{18}\right)+.. \\ $$$$\mathrm{amd}\:\mathrm{so}\:\mathrm{on}. \\ $$$$\mathrm{look}\:\mathrm{here}\:\mathrm{nothing}\:\mathrm{is}\:\mathrm{repeated}.\:\mathrm{You}\:\mathrm{have}\:\mathrm{to}\: \\ $$$$\mathrm{take}\:\mathrm{six}\:\mathrm{numbers}\:\mathrm{as}\:\mathrm{a}\:\mathrm{group}\:\mathrm{like} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}−\mathrm{4}−\mathrm{5}−\mathrm{6}\:\: \\ $$$$\mathrm{then} \\ $$$$\mathrm{7}+\mathrm{8}+\mathrm{9}−\mathrm{10}−\mathrm{11}−\mathrm{12} \\ $$$$\mathrm{then} \\ $$$$\mathrm{13}+\mathrm{14}+\mathrm{15}−\mathrm{16}−\mathrm{17}−\mathrm{18} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on}. \\ $$

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