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The sum of cubes of n natural numbers is a mathematical concept that finds the sum of the cubes of the first n natural numbers. Learn more about the sum of cubes of n natural numbers by reading below.
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Sum of cubes of n natural numbers
The sum of cubes of the first n natural numbers is a commonly studied problem in mathematics. It is often denoted by the symbol Σn³, where Σ is the summation notation and n is the number of terms being summed.
The formula for the sum of cubes of the first n natural numbers is:
Σn³ = 1³ + 2³ + 3³ + … + n³ = (n(n+1)/2)²
To derive this formula, we can use a method called mathematical induction. The idea behind mathematical induction is to prove that a statement is true for all positive integers by first showing that it is true for the smallest positive integer (usually 1), and then showing that if it is true for some integer k, then it must also be true for the next integer k+1.
To prove that the formula is true for n=1, we simply substitute n=1 into the formula and get:
Σ1³ = 1³ = 1 = (1(1+1)/2)²
This shows that the formula is true for n=1.
Next, we assume that the formula is true for some integer k, which means:
Σk³ = 1³ + 2³ + 3³ + … + k³ = (k(k+1)/2)²
Now, we want to show that the formula is also true for k+1, which means:
Σ(k+1)³ = 1³ + 2³ + 3³ + … + k³ + (k+1)³
We can rewrite this expression as:
Σ(k+1)³ = Σk³ + (k+1)³
Using the formula we assumed to be true for k, we can substitute in the value of Σk³:
Σ(k+1)³ = (k(k+1)/2)² + (k+1)³
Simplifying this expression gives:
Σ(k+1)³ = [(k+1)/2]² * [(2k(k+1))/2 + (k+1)]
Σ(k+1)³ = [(k+1)/2]² * [(2k² + 3k + 1)]
Σ(k+1)³ = [(k+1)/2]² * [(k+1)(2k + 1)]
Σ(k+1)³ = [(k+1)(k+2)/2]²
This expression is exactly the formula we started with, but with n replaced by k+1. Therefore, we have shown that if the formula is true for some integer k, then it must also be true for k+1. Since we have already shown that the formula is true for n=1, we can conclude that it is true for all positive integers.
In conclusion, the sum of cubes of the first n natural numbers is given by the formula Σn³ = (n(n+1)/2)². This formula can be derived using mathematical induction, which shows that the formula is true for all positive integers.
Sum of cubes of n natural numbers formula
The sum of cubes of the first n natural numbers is a mathematical problem that involves finding the sum of the cubes of the first n natural numbers, where n is a positive integer. This problem has a formula that can be used to quickly find the sum of the cubes of the first n natural numbers.
The formula for the sum of cubes of the first n natural numbers is:
Σn³ = 1³ + 2³ + 3³ + … + n³ = (n(n+1)/2)²
The formula is based on the fact that the sum of the first n natural numbers is given by the formula:
Σn = 1 + 2 + 3 + … + n = n(n+1)/2
To derive the formula for the sum of cubes of the first n natural numbers, we can use a method called mathematical induction. The idea behind mathematical induction is to prove that a statement is true for all positive integers by first showing that it is true for the smallest positive integer (usually 1), and then showing that if it is true for some integer k, then it must also be true for the next integer k+1.
To prove the formula for n=1, we can simply substitute n=1 into the formula and get:
Σ1³ = 1³ = 1 = (1(1+1)/2)²
This shows that the formula is true for n=1.
Next, we assume that the formula is true for some integer k, which means:
Σk³ = 1³ + 2³ + 3³ + … + k³ = (k(k+1)/2)²
Now, we want to show that the formula is also true for k+1, which means:
Σ(k+1)³ = 1³ + 2³ + 3³ + … + k³ + (k+1)³
We can rewrite this expression as:
Σ(k+1)³ = Σk³ + (k+1)³
Using the formula we assumed to be true for k, we can substitute in the value of Σk³:
Σ(k+1)³ = (k(k+1)/2)² + (k+1)³
After simplification, the expression becomes [(k+1)(k+2)/2]², which is the same as the original formula with n replaced by k+1. This shows that if the formula is valid for some integer k, it must also be valid for k+1. As we have already established that the formula holds true for n=1, we can deduce that it is true for all positive integers.
In conclusion, the formula for the sum of cubes of the first n natural numbers is Σn³ = (n(n+1)/2)². This formula can be derived using mathematical induction, which shows that the formula is true for all positive integers. The formula can be useful in many areas of mathematics and is often used in calculus and number theory.
What is the formula of sum of cubes of n natural numbers?
The formula for the sum of cubes of the first n natural numbers is an expression that can be used to quickly find the sum of the cubes of the first n natural numbers. The sum of cubes of the first n natural numbers is a mathematical problem that involves finding the sum of the cubes of the first n natural numbers, where n is a positive integer.
The formula for the sum of cubes of the first n natural numbers is given by:
Σn³ = 1³ + 2³ + 3³ + … + n³ = (n(n+1)/2)²
This formula is based on the fact that the sum of the first n natural numbers is given by the formula:
Σn = 1 + 2 + 3 + … + n = n(n+1)/2
To derive the formula for the sum of cubes of the first n natural numbers, we can use a method called mathematical induction. The idea behind mathematical induction is to prove that a statement is true for all positive integers by first showing that it is true for the smallest positive integer (usually 1), and then showing that if it is true for some integer k, then it must also be true for the next integer k+1.
To prove the formula for n=1, we can substitute n=1 into the formula and get:
Σ1³ = 1³ = 1 = (1(1+1)/2)²
This shows that the formula is true for n=1.
Next, we assume that the formula is true for some integer k, which means:
Σk³ = 1³ + 2³ + 3³ + … + k³ = (k(k+1)/2)²
Now, we want to show that the formula is also true for k+1, which means:
Σ(k+1)³ = 1³ + 2³ + 3³ + … + k³ + (k+1)³
We can rewrite this expression as:
Σ(k+1)³ = Σk³ + (k+1)³
Using the formula we assumed to be true for k, we can substitute in the value of Σk³:
Σ(k+1)³ = (k(k+1)/2)² + (k+1)³
Simplifying this expression gives:
Σ(k+1)³ = [(k+1)/2]² * [(2k(k+1))/2 + (k+1)]
Σ(k+1)³ = [(k+1)/2]² * [(2k² + 3k + 1)]
Σ(k+1)³ = [(k+1)/2]² * [(k+1)(2k + 1)]
Σ(k+1)³ = [(k+1)(k+2)/2]²
This expression is exactly the formula we started with, but with n replaced by k+1. Therefore, we have shown that if the formula is true for some integer k, then it must also be true for k+1. Since we have already shown that the formula is true for n=1, we can conclude that it is true for all positive integers.
Sum of cubes of n natural numbers calculator
The sum of the cubes of the first n natural numbers can be calculated using a formula. This formula is derived using mathematical induction, and it is given by:
sum of cubes of n natural numbers = [n(n+1)/2]^2
To understand this formula, let’s first consider the sum of the cubes of the first few natural numbers:
sum of cubes of first 1 natural number = 1^3 = 1
sum of cubes of first 2 natural numbers = 1^3 + 2^3 = 9
sum of cubes of first 3 natural numbers = 1^3 + 2^3 + 3^3 = 36
sum of cubes of first 4 natural numbers = 1^3 + 2^3 + 3^3 + 4^3 = 100
From the above examples, we can see that the sum of cubes of the first n natural numbers can be expressed as a sum of cubes of the first (n-1) natural numbers plus n^3. That is,
sum of cubes of first n natural numbers = sum of cubes of first (n-1) natural numbers + n^3
Using this relationship, we can prove the formula for the sum of cubes of the first n natural numbers using mathematical induction.
Base case: For n=1, the formula gives (1*(1+1)/2)^2 = 1^2 = 1, which is true.
Induction hypothesis: Assume that the formula is true for some arbitrary natural number k, i.e.,
sum of cubes of first k natural numbers = [k(k+1)/2]^2
Induction step: We need to show that the formula is true for k+1. That is,
sum of cubes of first (k+1) natural numbers = [(k+1)((k+1)+1)/2]^2
We can use the relationship we derived earlier to express the sum of cubes of the first (k+1) natural numbers as a sum of cubes of the first k natural numbers plus (k+1)^3. That is,
sum of cubes of first (k+1) natural numbers = [k(k+1)/2]^2 + (k+1)^3
Now, we can simplify this expression:
sum of cubes of first (k+1) natural numbers = [(k^2+k)/2]^2 + (k+1)^3
= (k^4+2k^3+3k^2+2k+1)/4 + (k+1)^3
= [(k+1)(k^3+4k^2+5k+4)]/4
= [(k+1)(k+2)/2]^2
Therefore, the formula holds for k+1 as well.
Thus, we have proved the formula for the sum of cubes of the first n natural numbers using mathematical induction. We can now use this formula to calculate the sum of cubes of any number of natural numbers.
What is the formula of average of cubes of n natural numbers?
The formula for the average of cubes of the first n natural numbers can be derived using the formula for the sum of cubes of the first n natural numbers, which is:
sum of cubes of first n natural numbers = [n(n+1)/2]^2
To find the average of cubes of the first n natural numbers, we need to divide the sum of cubes of the first n natural numbers by n. That is,
average of cubes of first n natural numbers = sum of cubes of first n natural numbers / n
Substituting the formula for the sum of cubes of the first n natural numbers, we get:
average of cubes of first n natural numbers = [n(n+1)/2]^2 / n
Now, we can simplify this expression. First, we can square the numerator and then cancel out a factor of n from the numerator and denominator. That is,
average of cubes of first n natural numbers = [(n+1)/2]^2 * n
Simplifying further, we get:
average of cubes of first n natural numbers = [n(n+1)/4]^2
Therefore, the formula for the average of cubes of the first n natural numbers is:
average of cubes of first n natural numbers = [n(n+1)/4]^2
Let’s verify this formula with a few examples:
For n=1, the average of cubes of the first n natural numbers is:
average of cubes of first 1 natural number = [1(1+1)/4]^2
= 1/4
= 0.25
For n=2, the average of cubes of the first n natural numbers is:
average of cubes of first 2 natural numbers = [2(2+1)/4]^2
= 3/4^2
= 0.1875
For n=3, the average of cubes of the first n natural numbers is:
average of cubes of first 3 natural numbers = [3(3+1)/4]^2
= 10.125
As we can see, the formula gives the correct values for the average of cubes of the first few natural numbers. We can use this formula to find the average of cubes of any number of natural numbers.
Sum of cubes of n natural numbers – FAQ
1. What is the sum of cubes of the first 10 natural numbers?
The sum of cubes of the first 10 natural numbers is 3025.
2. What is the formula for the sum of cubes of n natural numbers?
The formula for the sum of cubes of n natural numbers is [n(n+1)/2]^2.
3. What is the sum of cubes of the first 100 natural numbers?
The sum of cubes of the first 100 natural numbers is 25502500.
4. What is the sum of cubes of the first 50 even natural numbers?
The sum of cubes of the first 50 even natural numbers is 21262500.
5. What is the sum of cubes of the first 25 odd natural numbers?
The sum of cubes of the first 25 odd natural numbers is 105625.
6. What is the sum of cubes of the first n even natural numbers?
The sum of cubes of the first n even natural numbers is [n(n+1)]^2.
7. What is the sum of cubes of the first n odd natural numbers?
The sum of cubes of the first n odd natural numbers is n^2(n^2-1)/4.
8. What is the sum of cubes of the first 5 prime natural numbers?
The sum of cubes of the first 5 prime natural numbers is 372.
9. What is the sum of cubes of the first n natural numbers if n is odd?
The sum of cubes of the first n natural numbers if n is odd is n^2[(n-1)/2]^2.
10. What is the sum of cubes of the first n natural numbers if n is even?
The sum of cubes of the first n natural numbers if n is even is [n/2]^2(n+1)^2.
11. What is the sum of cubes of the first n natural numbers if n is a perfect square?
The sum of cubes of the first n natural numbers if n is a perfect square is [n(n+1)/2]^2.
12. What is the sum of cubes of the first n natural numbers if n is a perfect cube?
The sum of cubes of the first n natural numbers if n is a perfect cube is [n(n+1)/2]^2.
13. What is the sum of cubes of the first n natural numbers if n is a triangular number?
The sum of cubes of the first n natural numbers if n is a triangular number is [(n(n+1)/2)]^2.
14. What is the sum of cubes of the first n natural numbers if n is a pentagonal number?
The sum of cubes of the first n natural numbers if n is a pentagonal number is [n(3n-1)/2]^2.
15. What is the sum of cubes of the first n natural numbers if n is a hexagonal number?
The sum of cubes of the first n natural numbers if n is a hexagonal number is [n(2n-1)]^2.
16. What is the sum of cubes of the first n natural numbers if n is a perfect fifth power?
The sum of cubes of the first n natural numbers if n is a perfect fifth power is [n(n+1)/2]^2.
17. What is the sum of cubes of the first n natural numbers if n is a perfect sixth power?
The sum of cubes of the first n natural numbers if n is a perfect sixth power is [n(n+1)/2]^2.
18. What is the sum of cubes of the first n natural numbers if n is a Fibonacci number?
The sum of cubes of the first n natural numbers if n is a Fibonacci number is F(n)^2 * F(n+1)^2 / 4, where F(n) is the nth Fibonacci number.
19. What is the sum of cubes of the first n natural numbers if n is a Mersenne prime?
The sum of cubes of the first n natural numbers if n is a Mersenne prime is [2^(n-1)][2^(n)-1]^2.
20. What is the sum of cubes of the first n natural numbers if n is a perfect fourth power?
The sum of cubes of the first n natural numbers if n is a perfect fourth power is [n(n+1)(2n+1)(3n^2+3n-1)]/30.
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