The numbers a_n^ s are defined by a₀=1, a_ n+1 =3 n^2+n+a_n,(n 0) Then, a_n is equal to

Step-by-step solution

Given, a₀=1 and a_ n+1 =3 n^2+n+a_n a_n=3(n-1)^2+(n-1)+a_ n-1 Now, put n=1,2,3, , n We get, a₁=0+a₀=0+1=1 array ll & a₂=4+a₁ & a₃=14+a₂ & a₄=30+a₃ array ........ ........ ........ array ll & a_ n-1 =3(n-2)^2+(n-2)+a_ n-2 & a_n=3(n-1)^2+(n-1)+a_ n-1 array On adding, we get a_n= (1+4+14+30+ +3(n-1)^2+(n-1) ) array ll & a_n= 3 (n^2+1-2 n )+(n-1) & a_n= (3 n^2+3-6 n+n-1 ) & a_n= (3 n^2-5 n+2 ) & a_n=3 n^2-5 +2 1 array a_n= 3 n(n+1)(2 n+1) 6 - 5 n(n+1) 2 +2 n a_n= n 2 (n+1) 2 n+1-5 +2 n a_n= n 2 (n+1)(2 n-4)+2 n=n(n+1)(