The sum of 1 1/3 1/32 1/33 ……. 1/3n-1?
The given series is 1, 1/3, 1/32, 1/33, and so on. We need to find the sum of this series. Let's break down the problem into smaller steps to understand it better.
Step 1: Understanding the pattern
Looking at the series, we can observe that the denominator of each term starts from 3 and increases by 1 in each subsequent term. The numerator remains constant as 1 throughout the series.
Step 2: Writing the terms of the series
Let's write the first few terms of the series to get a better understanding:
1, 1/3, 1/32, 1/33, 1/34, 1/35, ...
Step 3: Finding a general term
To find the general term of the series, we need to identify a pattern. Looking closely, we can see that the denominator of each term is given by 3 + (n-1), where n represents the position of the term in the series. The numerator remains 1 throughout the series. Therefore, the general term can be written as 1/(3+(n-1)) or 1/(2+n).
Step 4: Finding the sum of the series
To find the sum of the series, we need to add up all the terms. Let's denote the sum of the series by S.
S = 1 + 1/3 + 1/32 + 1/33 + ...
Step 5: Simplifying the sum
We can simplify the sum by writing it as a single fraction. To do this, we need to find a common denominator for all the terms.
Step 6: Finding the least common denominator (LCD)
The least common denominator for the terms in the series is 3^2 * 2 * 33 * 34 * 35. This is because the denominator of each term can be expressed as a product of factors of 3 and factors of n.
Step 7: Writing all the terms with the LCD
Now, let's write all the terms of the series with the LCD as the denominator:
S = (3^2 * 2 * 33 * 34 * 35)/[(3^2 * 2 * 33 * 34 * 35)] + (3^2 * 2 * 33 * 34 * 35)/[(3^2 * 2 * 3 * 34 * 35)] + (3^2 * 2 * 33 * 34 * 35)/[(3^2 * 32 * 33 * 35)] + ...
Step 8: Adding all the terms
Since all the terms now have the same denominator, we can add them together:
S = [(3^2 * 2 * 33 * 34 * 35) + (3^2 * 2 * 33 * 34 * 35)/(3^2 * 2 * 3 * 34 * 35) + (3^2 * 2 * 33 * 34 * 35)/(3^2 * 32 * 33 * 35) + ...]
Step 9: Factoring out common terms