so 3n2 = 15 ie n2 = 5 so n = sqrt 5
t(n) = 3n2 + n = n(3n + 1)
(3n+2)(n+1)
3 6
The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6
so 3n2 = 15 ie n2 = 5 so n = sqrt 5
n3 + 3n2 + 4n + 12 = (n3 + 3n2) + (4n + 12) = n2(n + 3) + 4(n + 3) = (n2 + 4)(n + 3).
Add 7 to each side: 3n2 = 7 Divide each side by 3: n2 = 7/3 n = sqrt 7/sqrt 3, ie just over 1.5
-((3n - 1)(n + 3))
3n2 - 7n = 0 n(3n - 7) = 0 So either n = 0 or 3n-7 = 0 ie, either n = 0 or n = 7/3 = 21/3 = 2.33...
t(n) = 3n2 + 5n + 1 for n = 1, 2, 3, ...
Given m equals 3 and n equals 1 then m3n3 equals?m3n3 = m*3*n*3 = 3*3*1*3 = [ 27 ]
n = 3
t(n) = 3n2 + n = n(3n + 1)
(3n+2)(n+1)
30. The series can be generated by: t(n) = (-n3 + 3n2 - 56n + 510)/6 for n = 1, 2, 3, ...
3 6