What is the sum of the cubes of the first n natural numbers?
We know that sum of cubes of first n natural numbers is = (n(n+1)/2)2.
How do you find the sum of the cubes of n natural numbers?
Solution:
- We know the sum of the cubes of first n natural numbers (S) = {n(n+1)2}2.
- Therefore, the sum of the cubes of first 12 natural numbers = {12(12+1)2}2.
- We know the sum of the cubes of first n natural numbers (S) = {n(n+1)2}2.
- Therefore, the sum of the cubes of first 25 natural numbers = {25(25+1)2}2.
What is the formula to find sum of cubes?
The sum of cubes formula is one of the important algebraic identity. It is represented by a3 + b3 and is read as a cube plus b cube. The sum of cubes (a3 + b3) formula is expressed as a3 + b3 = (a + b) (a2 – ab + b2).
What is the sum of cubes of first four natural numbers?
+103=(10×11/2)2=552 = 3025.
What is the sum of squares of first n natural numbers?
Sum of Squares of n Natural Numbers Formula Natural numbers include whole numbers in them except the number 0. If we need to calculate the sum of squares of n consecutive natural numbers, the formula is Σn2 = n×(n+1)×(2n+1)6 n × ( n + 1 ) × ( 2 n + 1 ) 6 . It is easy to apply the formula when the value of n is known.
What is the sum of cubes of first 8 natural numbers?
1296
The sum of cubes of the first 8 natural numbers is 1296. Let us look at how we got the answer. Sum of cubes of 8 natural numbers = [82 (8 + 1)2]/4 = (64 × 81)/4 = 1296.
What is the sum of squares of first 10 natural numbers?
The sum of the squares of the first ten natural numbers is, 12 + 22 + + 102 = 385 The square of the sum of the first ten natural numbers is, (1 + 2 + + 10)2 = 552 = 3025 Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.