1st term= 3 2nd term = 5 Nth term = 2n+1 10th term= 21 = 2(10)+1
The nth term of the sequence is 2n + 1.
Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.
There is no solution to the question as asked. If the sum of n terms is 2n+1 then the sum of n+1 terms, using the same formula, is 2*(n+1)+1 = 2n+2+1 = 2n+3 So the (n+1)th term is sum to n+1 minus sum to n = (2n+3) - (2n+1) = 2 So each term is 2. But if each term is 2, then the sum of n terms must be even. The sum is clearly odd - which leads to a contradiction.
41.
2n-1 to the tenth term = 1
1st term= 3 2nd term = 5 Nth term = 2n+1 10th term= 21 = 2(10)+1
The nth term of the sequence is 2n + 1.
Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.
t(n) = 50 - 2n where n = 1, 2, 3, ...
There is no solution to the question as asked. If the sum of n terms is 2n+1 then the sum of n+1 terms, using the same formula, is 2*(n+1)+1 = 2n+2+1 = 2n+3 So the (n+1)th term is sum to n+1 minus sum to n = (2n+3) - (2n+1) = 2 So each term is 2. But if each term is 2, then the sum of n terms must be even. The sum is clearly odd - which leads to a contradiction.
-n2+2n+49
41.
2n-1
Let, n=10 2n+1=? 2(10)+1= 21
n = 1, 2n = 2 n = 2, 2n = 4 n = 3, 2n = 6 2, 4, 6, ..., 2n where n = 1, 2, 3, ... This is an arithmetic sequence, where the first term is 2 and the common difference is 2.
The nth term is: 5-2n