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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 .
Findthesumoftheseries13−23+33−43+…+(2n+1)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)
Thisisacompoundseries.13−23+33−43+…−(2n)3+(2n+1)3=13−{23+43…+(2n)3}+{33+53+…+(2n+1)3}=1−{2[n(n+1)]2}+{(13+33+…+(2n−1)3)−13+(2n+1)3}=1−2[n(n+1)]2+{n2(2n2−1)−1+(2n+1)3}=−2[n(n+1)]2+n2(2n2−1)+(2n+1)3=n2{2n2−1−2(n2+2n+1)}+(2n+1)3=n2{2n2−1−2n2−4n−2}+(2n+1)3=(2n+1)3−n2(4n+3)=4n3+9n2+6n+1Verification:LetS=13−23+33−43+53=81(Directly)Byformula:(2n+1)3=53⇒n=2S=4(2)3+9(2)2+6(2)+1=4(8)+9(4)+6(2)+1=32+36+12+1=81(Same)Formulaeusedinabove23+43+63+…+(2n)3=2[n(n+1)]213+33+53+…+(2n−1)3=n2(2n2−1)

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