how is this series 3+ 9/4 +27/16 +81/64... is converge to 12?
i represent the series as 3^(n+1)/4^n n starts from 0 to infinity, is this correct formula
As n gets larger the sum goes to 0
To determine if the given series converges to 12, we need to find its sum by using the formula for the sum of a geometric series:
S = a / (1 - r)
In this case, the first term (a) is 3, and the common ratio (r) is 3/4. Let's plug these values into the formula:
S = 3 / (1 - 3/4)
Simplifying the denominator, we have:
S = 3 / (4/4 - 3/4)
= 3 / (1/4)
= 12
Therefore, the sum of the series 3 + 9/4 + 27/16 + 81/64... converges to 12.
To determine whether the series 3 + 9/4 + 27/16 + 81/64... converges and finds its sum, we can use the formula for the sum of an infinite geometric series.
In this case, the series can be written as:
S = 3 + 9/4 + 27/16 + 81/64...
To determine if the series converges, we need to find the common ratio (r) between consecutive terms. Looking at the series, we can see that each term is obtained by multiplying the previous term by 3/4. So, we can write the common ratio as:
r = 3/4
Now, using the formula for the sum of an infinite geometric series, which is given by:
S = a / (1 - r)
where:
S = sum of the series
a = first term of the series
r = common ratio
Substituting the values into the formula:
S = 3 / (1 - 3/4)
Simplifying further:
S = 3 / (1/4)
To divide by a fraction, we can multiply it by the reciprocal:
S = 3 * (4/1)
S = 12
Hence, the sum of the series is 12.
3*[1 + 3/4 + (3/4)^2 + ...(3/4)^n]
As n-> infnity, this becomes
3*[1/(1 - (3/4)] = 3*4 = 12