5016. Here's how:
Each term is 1 less than the terms in 9, 18, 27 ..., which is the sequence 9*N, with N = 1..33.
So the actual formula for the sequence is {9*N - 1}
The sequence 1,2,3...N can be summed by (N+1)*N/2 --> N=33: (33+1)*33/2 = 561, so the sum of the first 33 terms of 9+18+27+297 = 9*(1+2+3+..+33) = 9*561 = 5049. But our sequence terms are each one less than the terms in 9,18,27, so subtract 33 from 5049 = 5016.
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
Sum of 1st 2 terms, A2 = 2 + 4 = 6 Sum of 1st 3 terms, A3 = 2 + 4 + 6 = 12 Sum of 1st 4 terms A4 = 2 + 4 + 6 + 12 = 20 you can create a formula for the sum of the 1st n terms of this sequence Sum of 1st n terms of this sequence = n2 + n so the sum of the first 48 terms of the sequence is 482 + 48 = 2352
A binary sequence is a sequence of [pseudo-]randomly generated binary digits. There is no definitive sum because the numbers are random. The sum could range from 0 to 64 with a mean sum of 32.
49
Find the sum of the first 11 terms in the sequence 3 7 11
To find the sum of the first 28 terms of an arithmetic sequence, you need the first term (a) and the common difference (d). The formula for the sum of the first n terms (S_n) of an arithmetic sequence is S_n = n/2 * (2a + (n - 1)d). Once you have the values of a and d, plug them into the formula along with n = 28 to calculate the sum.
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
Sum of 1st 2 terms, A2 = 2 + 4 = 6 Sum of 1st 3 terms, A3 = 2 + 4 + 6 = 12 Sum of 1st 4 terms A4 = 2 + 4 + 6 + 12 = 20 you can create a formula for the sum of the 1st n terms of this sequence Sum of 1st n terms of this sequence = n2 + n so the sum of the first 48 terms of the sequence is 482 + 48 = 2352
The terms of a sequence added together is the sum.
A binary sequence is a sequence of [pseudo-]randomly generated binary digits. There is no definitive sum because the numbers are random. The sum could range from 0 to 64 with a mean sum of 32.
3925
because you add the first 2 terms and the next tern was the the sum of the first 2 terms.
49
Find the sum of the first 11 terms in the sequence 3 7 11
The sum of the first 12 terms of an arithmetic sequence is: sum = (n/2)(2a + (n - 1)d) = (12/2)(2a + (12 - 1)d) = 6(2a + 11d) = 12a + 66d where a is the first term and d is the common difference.
nth term is 8 - n. an = 8 - n, so the sequence is {7, 6, 5, 4, 3, 2,...} (this is a decreasing sequence since the successor term is smaller than the nth term). So, the sum of first six terms of the sequence is 27.
yup