1st term= 3 2nd term = 5 Nth term = 2n+1 10th term= 21 = 2(10)+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.
That factors to (2n + 1)(n + 2)
(4n + 1)*(2n + 6) = 8n2 + 26n + 6
6=(1-2n)+5 6-5=(1-2n)+5-5 1=1-2n 1-1=1-1-2n 0=-2n 0/-2=-2n/-2 0=n
1st term= 3 2nd term = 5 Nth term = 2n+1 10th term= 21 = 2(10)+1
2n-1 to the tenth term = 1
2n+4/2 term 1 = 3 term 2 = 4 term 3 = 5 term 4 = 6
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.
That factors to (2n + 1)(n + 2)
5n + 1 = 10 + 2n5n - 2n + 1 - 1 = 10 - 1 + 2n - 2n3n = 93n/3 = 9/3n = 3
(4n + 1)*(2n + 6) = 8n2 + 26n + 6
When n=53, then 2n + 1 = 107
6=(1-2n)+5 6-5=(1-2n)+5-5 1=1-2n 1-1=1-1-2n 0=-2n 0/-2=-2n/-2 0=n
The sum of the numbers given is 31. 1 + 2 + 4 + 8 + ... + 2n = 2n+1 - 1
aneuploidy
The nth term of the sequence is 2n + 1.