find the sum of the infinite geometric series
1/3+1/9+1/27
sum=a_1/1-r
please help me.
well, clearly
a_1 = 1/3
and r = 1/3
so just plug in your values:
S = a/(1-r) = (1/3)/(1 - 1/3) = 1/2
To find the sum of an infinite geometric series, you can use the formula:
Sum = a_1 / (1 - r),
where "a_1" is the first term of the series and "r" is the common ratio between consecutive terms.
In your case, the first term (a_1) is 1/3 and the common ratio (r) is 1/3 as well as each term is obtained by multiplying the previous term by 1/3.
Now, let's substitute the values into the formula:
Sum = (1/3) / (1 - 1/3)
To simplify the expression, we need to find a common denominator for the fractions in the denominator:
Sum = (1/3) / (3/3 - 1/3)
Sum = (1/3) / (2/3)
To divide fractions, we can multiply the first fraction by the reciprocal of the second fraction:
Sum = (1/3) * (3/2)
Now we can cancel out the common factor of 3:
Sum = (1/1) * (1/2)
Finally, we find:
Sum = 1/2
Therefore, the sum of the infinite geometric series 1/3 + 1/9 + 1/27 + ... is 1/2.