(3n+2)(n+1)
t(n) = 5n - 3
1st term= 3 2nd term = 5 Nth term = 2n+1 10th term= 21 = 2(10)+1
That factors to (2n + 1)(n + 2)
5n + 3 = 14 - 6n <=> 5n +6n = 14 - 3 <=> 11n = 11 => n = 11/11 = 1
4 over 5n
(2n+3)(n+1)=2n2+5n+3.
9 - 5n + 1 = 15 -5n = 15 - 10 n = -5/5 n = -1
5n + 1 = 10 + 2n5n - 2n + 1 - 1 = 10 - 1 + 2n - 2n3n = 93n/3 = 9/3n = 3
(3n+2)(n+1)
t(n) = 5n - 3
It is 1 + 5N.
1st term= 3 2nd term = 5 Nth term = 2n+1 10th term= 21 = 2(10)+1
9-5n+1 = 15 -5n = 15-9-1 -5n = 5 Divide both sides of the equation by -5 to find the value of n: n = -5
(5n + 1)(n + 7)
That factors to (2n + 1)(n + 2)
5n + 3 = 14 - 6n <=> 5n +6n = 14 - 3 <=> 11n = 11 => n = 11/11 = 1