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

Commented bymr W last updated on 23/Jul/20

Commented bymr 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

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