If the nth term of a sequence is u1,u2,u3...,is 2n-1,find the sum of it's first eight terms
To find the sum of the first eight terms of the sequence, we need to substitute n from 1 to 8 in the formula 2n-1 and then add up all the values.
1st term: 2(1) - 1 = 1
2nd term: 2(2) - 1 = 3
3rd term: 2(3) - 1 = 5
4th term: 2(4) - 1 = 7
5th term: 2(5) - 1 = 9
6th term: 2(6) - 1 = 11
7th term: 2(7) - 1 = 13
8th term: 2(8) - 1 = 15
Now we add up all these terms:
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64
Therefore, the sum of the first eight terms of the sequence is 64.
This sure looks like an arithmetic sequence Tn = a + (n-1)d
Tn = 1 + (n-1) 2
in which case the sum of 8 is'
(n/2) [ 2 a + (n-1)d ] = (8/2) [ 2 + 7*2 ] = 4 4 *16 = 64 sure enough
Yes, you are correct. The sequence given indeed follows the form of an arithmetic sequence, where the nth term is given by Tn = a + (n-1)d.
By using the formula for the sum of the first n terms of an arithmetic sequence, which is Sn = (n/2)[2a + (n-1)d], we can calculate the sum of the first eight terms as follows:
Sn = (8/2)[2*1 + (8-1)2]
= 4[2 + 7*2]
= 4[2 + 14]
= 4 * 16
= 64
Therefore, the sum of the first eight terms of the sequence is indeed 64. Thank you for pointing out the arithmetic sequence pattern in the sequence.