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Question Number 1784 by 112358 last updated on 25/Sep/15
Find the sum of the series   1^3 −2^3 +3^3 −4^3 +...+(2n+1)^3  .
$${Find}\:{the}\:{sum}\:{of}\:{the}\:{series}\: \\ $$$$\mathrm{1}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} −\mathrm{4}^{\mathrm{3}} +…+\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \:. \\ $$
Answered by Rasheed Soomro last updated on 25/Sep/15
This is a compound series.  1^3 −2^3 +3^3 −4^3 +...−(2n)^3 +(2n+1)^3         =1^3 −{2^3 +4^3 ...+(2n)^3  }+{3^3 +5^3 +...+(2n+1)^3 }       =1−{2[n(n+1)]^2 }+{(1^3 +3^3 +...+(2n−1)^3 )−1^3 +(2n+1)^3 }      =1−2[n(n+1)]^2 +{n^2 (2n^2 −1)−1+(2n+1)^3 }      =−2[n(n+1)]^2 +n^2 (2n^2 −1)+(2n+1)^3       =n^2 {2n^2 −1−2(n^2 +2n+1)}+(2n+1)^3       =n^2 {2n^2 −1−2n^2 −4n−2}+(2n+1)^3       =(2n+1)^3 −n^2 (4n+3)      =4n^3 +9n^2 +6n+1  Verification:  Let   S=1^3 −2^3 +3^3 −4^3 +5^3 =81 (Directly)  By formula:  (2n+1)^3 =5^3 ⇒n=2  S=4(2)^3 +9(2)^2 +6(2)+1     =4(8)+9(4)+6(2)+1     =32+36+12+1=81 (Same)  Formulae used in above         2^3 +4^3 +6^3 +...+(2n)^3 =2[n(n+1)]^2          1^3 +3^3 +5^3 +...+(2n−1)^3 =n^2 (2n^2 −1)
$${This}\:{is}\:{a}\:{compound}\:{series}. \\ $$$$\mathrm{1}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} −\mathrm{4}^{\mathrm{3}} +…−\left(\mathrm{2}{n}\right)^{\mathrm{3}} +\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \: \\ $$$$\:\:\:\:\:=\mathrm{1}^{\mathrm{3}} −\left\{\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} …+\left(\mathrm{2}{n}\right)^{\mathrm{3}} \:\right\}+\left\{\mathrm{3}^{\mathrm{3}} +\mathrm{5}^{\mathrm{3}} +…+\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \right\} \\ $$$$\:\:\:\:\:=\mathrm{1}−\left\{\mathrm{2}\left[{n}\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}} \right\}+\left\{\left(\mathrm{1}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} \right)−\mathrm{1}^{\mathrm{3}} +\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \right\} \\ $$$$\:\:\:\:=\mathrm{1}−\mathrm{2}\left[{n}\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}} +\left\{{n}^{\mathrm{2}} \left(\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{1}+\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \right\} \\ $$$$\:\:\:\:=−\mathrm{2}\left[{n}\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}} +{n}^{\mathrm{2}} \left(\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}\right)+\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:={n}^{\mathrm{2}} \left\{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}−\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)\right\}+\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:={n}^{\mathrm{2}} \left\{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}−\mathrm{2}{n}^{\mathrm{2}} −\mathrm{4}{n}−\mathrm{2}\right\}+\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:=\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} −{n}^{\mathrm{2}} \left(\mathrm{4}{n}+\mathrm{3}\right) \\ $$$$\:\:\:\:=\mathrm{4}{n}^{\mathrm{3}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{1} \\ $$$${Verification}: \\ $$$${Let}\:\:\:{S}=\mathrm{1}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} −\mathrm{4}^{\mathrm{3}} +\mathrm{5}^{\mathrm{3}} =\mathrm{81}\:\left({Directly}\right) \\ $$$${By}\:{formula}: \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{5}^{\mathrm{3}} \Rightarrow{n}=\mathrm{2} \\ $$$${S}=\mathrm{4}\left(\mathrm{2}\right)^{\mathrm{3}} +\mathrm{9}\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}\left(\mathrm{2}\right)+\mathrm{1} \\ $$$$\:\:\:=\mathrm{4}\left(\mathrm{8}\right)+\mathrm{9}\left(\mathrm{4}\right)+\mathrm{6}\left(\mathrm{2}\right)+\mathrm{1} \\ $$$$\:\:\:=\mathrm{32}+\mathrm{36}+\mathrm{12}+\mathrm{1}=\mathrm{81}\:\left({Same}\right) \\ $$$${Formulae}\:{used}\:{in}\:{above} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +…+\left(\mathrm{2}{n}\right)^{\mathrm{3}} =\mathrm{2}\left[{n}\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{1}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\mathrm{5}^{\mathrm{3}} +…+\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} ={n}^{\mathrm{2}} \left(\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$ \\ $$

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