Using the Summation formula for this sequence what is S12 for 4+8+16+...
I don’t know how to do this at all
Thanks in advance !
geometric sequence
a = 4
r = 2
https://www.mathsisfun.com/algebra/sequences-sums-geometric.html
To find the sum of the series 4 + 8 + 16 + ..., we can use the formula for the sum of a geometric series. The formula is given by:
S = a * (1 - r^n) / (1 - r)
where:
S = sum of the series
a = first term of the series
r = common ratio
n = number of terms
In our case, the first term 'a' is 4 and the common ratio 'r' is 2 (since each term is twice the previous term). We want to find the sum up to the 12th term, so 'n' is 12.
Plugging these values into the formula, we have:
S12 = 4 * (1 - 2^12) / (1 - 2)
Simplifying further:
S12 = 4 * (1 - 4096) / (1 - 2)
We can simplify the numerator:
S12 = 4 * (-4095) / (1 - 2)
S12 = -16,380
Therefore, the sum S12 of the series 4 + 8 + 16 + ... up to the 12th term is -16,380.
No problem! I can help you with that. To find the sum of an arithmetic sequence using the Summation formula, you need to know the first term (a), the common difference (d), and the number of terms (n). In this case, the first term is 4 and the common difference is 8 - 4 = 4.
To find the number of terms (n), you need to find the relationship between the terms of the sequence. In this sequence, you can see that each term is obtained by multiplying the previous term by 2. So, we can write the nth term of the sequence as a geometric sequence: a(n) = a * r^(n-1), where a is the first term and r is the common ratio. In this case, a = 4 and r = 2.
Now we can find the number of terms by solving the equation a(n) = 4 * 2^(n-1) > 12. We can start by trial and error until we find the smallest value of n that satisfies the equation.
By substituting values for n, we have:
For n = 1, a(n) = 4 * 2^(1-1) = 4 * 2^0 = 4.
For n = 2, a(n) = 4 * 2^(2-1) = 4 * 2^1 = 8.
For n = 3, a(n) = 4 * 2^(3-1) = 4 * 2^2 = 16.
So, the smallest value of n that satisfies the equation is 3. Hence, n = 3 will be our upper limit, and we want the sum of the first 12 terms.
Now, we can use the Summation formula to find the sum of the sequence. The formula is S(n) = (n/2) * (2a + (n-1)d), where S(n) is the sum of the first 'n' terms. In this case, we want to find S(12), so n = 12, a = 4, and d = 4.
By substituting these values into the formula, we have:
S(12) = (12/2) * (2(4) + (12-1)(4))
= 6 * (8 + 11(4))
= 6 * (8 + 44)
= 6 * 52
= 312.
Therefore, the sum of the first 12 terms of the sequence 4+8+16+... is 312.