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.
Formula: (Hg2)3N2
2NaN ---> 2Na + 3N2
The GCF is 3n.
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.
-((3n - 1)(n + 3))
Maybe: [Cd(NH3)5]3N2