Question Number 115095 by aurpeyz last updated on 23/Sep/20

Answered by Dwaipayan Shikari last updated on 23/Sep/20

Term nth=6n−1  Σ_(n=1) ^n 6n−1=3n(n+1)−n  3n(n+1)−n>1000  3n^2 +2n−1000>0  n^2 +((2n)/3)−((1000)/3)>0  (n+(1/3))^2 −(1/9)−((1000)/3)>0  (n+(1/3))^2 >((3001)/9)  n>−(1/3)+(√((3001)/9))  n>17.92  So n has to be 18

Commented byaurpeyz last updated on 23/Sep/20

pls what is 3n(n+1)−n?

Commented byDwaipayan Shikari last updated on 23/Sep/20

Σ_(n=1) ^n 6n−1  =6.((n(n+1))/2)−n

Answered by Aziztisffola last updated on 23/Sep/20

u_0 =5  & u_(n+1) =u_n +6  ⇒u_n =6n+5  Σ_(k=0) ^n u_k =((u_0 +u_n )/2)(n+1)>1000  (3n+5)(n+1)>1000  3n^2 +8n−995>0  solve for n∈N  ⇒n>((√(3001))/2)−(4/3)=17.....  ⇒n=18