Answer :
To find the sum of the first 6 terms of the series 11, 22, 44, ..., we need to recognize that this is a geometric series.
A geometric series can be defined by the first term and the common ratio. For this series:
- The first term [tex]\(a\)[/tex] is 11.
- The common ratio [tex]\(r\)[/tex] is 2 (since each term is twice the previous term).
The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
Here, [tex]\( a = 11 \)[/tex], [tex]\( r = 2 \)[/tex], and [tex]\( n = 6 \)[/tex].
Plugging these values into the formula, we get:
[tex]\[ S_6 = 11 \frac{2^6 - 1}{2 - 1} \][/tex]
First, calculate [tex]\( 2^6 \)[/tex]:
[tex]\[ 2^6 = 64 \][/tex]
Then use this result in the formula:
[tex]\[ S_6 = 11 \frac{64 - 1}{1} \][/tex]
[tex]\[ S_6 = 11 \frac{63}{1} \][/tex]
[tex]\[ S_6 = 11 \times 63 \][/tex]
[tex]\[ S_6 = 693 \][/tex]
So, the sum of the first 6 terms is 693.
Since 693 is already an integer, rounding to the nearest integer is not needed. Therefore, the sum of the first 6 terms of the given series, to the nearest integer, is:
[tex]\[ 693 \][/tex]
A geometric series can be defined by the first term and the common ratio. For this series:
- The first term [tex]\(a\)[/tex] is 11.
- The common ratio [tex]\(r\)[/tex] is 2 (since each term is twice the previous term).
The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
Here, [tex]\( a = 11 \)[/tex], [tex]\( r = 2 \)[/tex], and [tex]\( n = 6 \)[/tex].
Plugging these values into the formula, we get:
[tex]\[ S_6 = 11 \frac{2^6 - 1}{2 - 1} \][/tex]
First, calculate [tex]\( 2^6 \)[/tex]:
[tex]\[ 2^6 = 64 \][/tex]
Then use this result in the formula:
[tex]\[ S_6 = 11 \frac{64 - 1}{1} \][/tex]
[tex]\[ S_6 = 11 \frac{63}{1} \][/tex]
[tex]\[ S_6 = 11 \times 63 \][/tex]
[tex]\[ S_6 = 693 \][/tex]
So, the sum of the first 6 terms is 693.
Since 693 is already an integer, rounding to the nearest integer is not needed. Therefore, the sum of the first 6 terms of the given series, to the nearest integer, is:
[tex]\[ 693 \][/tex]