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.
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
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
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.
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
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
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.
yup
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.
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