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.
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
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.
Expressed in decimal, the sum of the numbers 1 to 8192 is 33558528 - expressed in binary, this number is equal to 10000000000001000000000000.
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....
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
The terms of a sequence added together is the sum.
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