What is the 2nd term of 3n-2?
Sn = 3n2 + 2n - 8
t(n) = 3n2 + n = n(3n + 1)
so 3n2 = 15 ie n2 = 5 so n = sqrt 5
(3n+2)(n+1)
The product is(the product of the first term of each)plus(the product of the last term of each) plus(the product of the first term of the first and the last term of the second) plus(the product of the first term of the second and the last term of the first).
Sn = 3n2 + 2n - 8
t(n) = 3n2 + n = n(3n + 1)
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).
3
n1 = 3n2 = 3(3) + 2 = 9 + 2 = 11n3 = 3(11) +2 = 33 + 2 = 35n4 = 3(35) + 2 = 105 + 2 = 107The fourth term in the sequence (n4) is 107.
The GCF is 3n.
Formula: (Hg2)3N2
2NaN ---> 2Na + 3N2
-((3n - 1)(n + 3))
The numbers are generated by Pn = (3n2 - n)/2 for n = 1, 2, 3, ... A list of pentagonal numbers (first 46) is given on the link below.
Maybe: [Cd(NH3)5]3N2