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Question Number 104706 by Skabetix last updated on 23/Jul/20
find the value of p (pls help) 1^3 +3^3 +5^3 +...+p^3 =8128
findthevalueofp(plshelp)13+33+53+…+p3=8128
Commented by mr W last updated on 23/Jul/20
Commented by mr W last updated on 23/Jul/20
n^2 (2n^2 −1)=8128 2n^4 −n^2 −8128=0 n^2 =((1+(√(1+4×2×8128)))/4)=64 n=8 p=2n−1=2×8−1=15
n2(2n2−1)=81282n4−n2−8128=0n2=1+1+4×2×81284=64n=8p=2n−1=2×8−1=15
Answered by Dwaipayan Shikari last updated on 23/Jul/20
T_n =(2n−1)^3 =8n^3 −1−12n^2 +6n ΣT_n =2(n(n+1))^2 −2n(n+1)(2n+1)+3n(n+1)−n =2n(n+1)(n^2 +n−2n−1)+n(3n+2) =n((2n+2)(n^2 −n−1))+n(3n+2) =n(2n^3 +2n^2 −2n^2 −2n−2n−2+3n+2) =n(2n^3 −n)=n^2 (2n^2 −1) n^2 (2n^2 −1)=8128 ⇒a(2a−1)=8128 ⇒2a^2 −a−8128=0 a=64 n=8 T_n =(2n−1)=15
Tn=(2n−1)3=8n3−1−12n2+6nΣTn=2(n(n+1))2−2n(n+1)(2n+1)+3n(n+1)−n=2n(n+1)(n2+n−2n−1)+n(3n+2)=n((2n+2)(n2−n−1))+n(3n+2)=n(2n3+2n2−2n2−2n−2n−2+3n+2)=n(2n3−n)=n2(2n2−1)n2(2n2−1)=8128⇒a(2a−1)=8128⇒2a2−a−8128=0a=64n=8Tn=(2n−1)=15

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