1. Question: Prove that the sum of an infinite geometric series with a common ratio less than 1 is finite.
Answer: To prove this, we can start by using the formula for the sum of an infinite geometric series, which is given by S = a / (1 – r), where ‘a’ is the first term and ‘r’ is the common ratio. Since the common ratio is less than 1, we can see that as the number of terms increases, the value of ‘r’ raised to the power of ‘n’ (where ‘n’ is the number of terms) approaches zero. Therefore, the sum of the series is finite.
2. Question: Find the sum of the series 1 + 3 + 5 + 7 + … + (2n – 1).
Answer: To find the sum of this series, we can use the formula for the sum of an arithmetic series, which is given by S = (n/2)(first term + last term). In this case, the first term is 1 and the last term is (2n – 1), so the sum of the series is S = (n/2)(1 + (2n – 1)). Simplifying this expression gives S = n^2.
3. Question: Prove that the sum of the squares of the first ‘n’ natural numbers is given by the formula n(n + 1)(2n + 1) / 6.
Answer: To prove this formula, we can use mathematical induction. First, we can show that the formula holds for n = 1. The sum of the squares of the first natural number is 1, and plugging n = 1 into the formula gives (1)(2)(3) / 6 = 1, which is true.
Next, we assume that the formula holds for some arbitrary value of ‘k’, and prove that it holds for ‘k + 1’. So, assuming the formula holds for k, we have the sum of squares of the first ‘k’ natural numbers given by k(k + 1)(2k + 1) / 6.
To prove that the formula holds for ‘k + 1’, we add (k + 1)^2 to the sum of squares of the first ‘k’ natural numbers. This gives us the sum of squares of the first ‘k + 1’ natural numbers, which can be simplified to (k + 1)(k + 2)(2k + 3) / 6.
By substituting the assumed formula for ‘k’, we can simplify this expression to (k + 1)(k + 2)(2k + 3) / 6 = (k + 1)(k + 2)(2k + 1) / 6 + (k + 1)^2.
Simplifying further, we get (k + 1)(k + 2)(2k + 1 + 6(k + 1)) / 6 = (k + 1)(k + 2)(2k + 1 + 6k + 6) / 6 = (k + 1)(k + 2)(8k + 7) / 6.
This expression matches the formula for the sum of squares of the first ‘k + 1’ natural numbers, so the formula holds for all natural numbers.
4. Question: Find the sum of the series 1/1! + 1/2! + 1/3! + … + 1/n!
Answer: To find the sum of this series, we can use the concept of the exponential function. The series can be written as e – 1, where e is the base of the natural logarithm. Therefore, the sum of the series is e – 1.
5. Question: Prove that the sum of the cubes of the first ‘n’ natural numbers is equal to the square of the sum of those numbers.
Answer: To prove this, we can start by using the formula for the sum of the first ‘n’ natural numbers, which is given by S = n(n + 1) / 2.
The sum of the cubes of the first ‘n’ natural numbers can be written as (1^3 + 2^3 + 3^3 + … + n^3). By using the formula for the sum of cubes, which is given by S = (n(n + 1) / 2)^2, we can see that this expression is equal to the square of the sum of the first ‘n’ natural numbers.
6. Question: Find the sum of the series 2 + 5 + 10 + 17 + … + (n^2 + 1).
Answer: To find the sum of this series, we can observe that each term is given by (n^2 + 1), where ‘n’ is the position of the term. By expanding this expression, we get n^2 + 1 = n^2 + n^0.
Therefore, the sum of the series can be written as (1^2 + 2^2 + 3^2 + … + n^2) + (1 + 1 + 1 + … + 1), where the first part is the sum of squares of the first ‘n’ natural numbers and the second part is the sum of ‘n’ ones.
Using the formula for the sum of squares, we know that the sum of squares of the first ‘n’ natural numbers is given by n(n + 1)(2n + 1) / 6. The sum of ‘n’ ones is simply n.
Therefore, the sum of the series is n(n + 1)(2n + 1) / 6 + n.
7. Question: Prove that the sum of the reciprocals of the Fibonacci sequence converges to a finite value.
Answer: To prove this, we can start by considering the ratio of consecutive Fibonacci numbers. As the sequence progresses, this ratio approaches the golden ratio, which is approximately 1.618.
Using this information, we can write the sum of the reciprocals of the Fibonacci sequence as S = 1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + …
By grouping terms, we can rewrite this expression as S = (1 + 1) + (1/2 + 1/3) + (1/5 + 1/8) + …
Simplifying further, we get S = 2 + (2/3) + (3/8) + …
Since the ratio of consecutive Fibonacci numbers approaches the golden ratio, we can see that the terms in the series approach 1/2, 1/3, 1/5, 1/8, …
Therefore, we can conclude that the sum of the reciprocals of the Fibonacci sequence converges to a finite value.
8. Question: Find the sum of the series (1/3) + (1/6) + (1/9) + … + (1/(3n)).
Answer: To find the sum of this series, we can observe that each term is given by 1/(3n), where ‘n’ is the position of the term. By simplifying this expression, we get 1/(3n) = 1/3 * (1/n).
Therefore, the sum of the series can be written as (1/3) * (1/1 + 1/2 + 1/3 + … + 1/n), where the part in parentheses is the sum of the reciprocals of the first ‘n’ natural numbers.
Using the formula for the sum of the reciprocals of the first ‘n’ natural numbers, which is given by (n(n + 1)) / 2n, we can simplify the expression to (1/3) * (n(n + 1)) / 2n.
Simplifying further, we get (n + 1) / 6.
Therefore, the sum of the series is (n + 1) / 6.
9. Question: Prove that the sum of the squares of the Fibonacci sequence is given by the product of the last two terms.
Answer: To prove this, we can start by considering the relationship between consecutive Fibonacci numbers. The ratio of consecutive Fibonacci numbers approaches the golden ratio, which is approximately 1.618.
Using this information, we can write the sum of the squares of the Fibonacci sequence as S = 1^2 + 1^2 + 2^2 + 3^2 + 5^2 + 8^2 + …
By grouping terms, we can rewrite this expression as S = (1 + 1) + (2^2 + 3^2) + (5^2 + 8^2) + …
Simplifying further, we get S = 2 + 13 + 89 + …
Since the ratio of consecutive Fibonacci numbers approaches the golden ratio, we can see that the terms in the series approach 2, 13, 89, …
Therefore, we can conclude that the sum of the squares of the Fibonacci sequence is given by the product of the last two terms.
10. Question: Find the sum of the series (1/2) + (1/4) + (1/8) + … + (1/2^n).
Answer: To find the sum of this series, we can observe that each term is given by 1/(2^n), where ‘n’ is the position of the term. By simplifying this expression, we get 1/(2^n) = 1/2 * (1/2)^(n-1).
Therefore, the sum of the series can be written as (1/2) * (1 + 1/2 + 1/4 + … + 1/2^(n-1)), where the part in parentheses is the sum of the geometric sequence with a common ratio of 1/2.
Using the formula for the sum of a geometric series, which is given by S = a / (1 – r), where ‘a’ is the first term and ‘r’ is the common ratio, we can simplify the expression to (1/2) * (1 / (1 – 1/2)).
Simplifying further, we get (1/2) * (1 / (1/2)) = (1/2) * 2 = 1.
Therefore, the sum of the series is 1.