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Sum of cubes of first n natural numbers Solution

STEP 0: Pre-Calculation Summary
Formula Used
sum_of_first_n_terms = ((Value of n*(Value of n+1))^2)/4
Sn = ((n*(n+1))^2)/4
This formula uses 1 Variables
Variables Used
Value of n- Value of n is the index value of position n in a series or a sequence.
STEP 1: Convert Input(s) to Base Unit
Value of n: 3 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
Sn = ((n*(n+1))^2)/4 --> ((3*(3+1))^2)/4
Evaluating ... ...
Sn = 36
STEP 3: Convert Result to Output's Unit
36 --> No Conversion Required
FINAL ANSWER
36 <-- Sum of First n terms
(Calculation completed in 00.000 seconds)

9 General Series Calculators

Sum of n natural numbers taken power of four
sum_of_n_natural_numbers_taken_power_of_4 = (Total terms*(Total terms+1)*((6*Total terms^3)+(9*Total terms^2)+Total terms-1))/30 Go
Sum of squares of first n even numbers
sum_of_square_of_first_n_even_number = (2*Value of n*(Value of n+1)*(2*Value of n+1))/3 Go
Sum of squares first n odd numbers
sum_of_square_of_first_n_odd_number = (Value of n*(2*Value of n+1)*(2*Value of n-1))/3 Go
Sum of squares of first n natural numbers
sum_of_first_n_terms = (Value of n*(Value of n+1)*(2*Value of n+1))/6 Go
Sum of cubes of first n natural numbers
sum_of_first_n_terms = ((Value of n*(Value of n+1))^2)/4 Go
Sum of first n natural numbers
sum_of_first_n_terms = (Value of n*(Value of n+1))/2 Go
Sum of first n even natural numbers
sum_of_first_n_terms = (Value of n*(Value of n+1)) Go
Sum of cubes of first n even numbers
sum_required = 2*(Value of n*(Value of n+1))^2 Go
Sum of first n odd natural numbers
sum_of_first_n_terms = (Value of n)^2 Go

Sum of cubes of first n natural numbers Formula

sum_of_first_n_terms = ((Value of n*(Value of n+1))^2)/4
Sn = ((n*(n+1))^2)/4

What does cubes of first n natural numbers mean?

Cubes of first n natural numbers simply means natural numbers raised to a power of 3. e.g- some of the cubes of starting natural numbers are 1,8,27,64,125 & so on.

How to Calculate Sum of cubes of first n natural numbers?

Sum of cubes of first n natural numbers calculator uses sum_of_first_n_terms = ((Value of n*(Value of n+1))^2)/4 to calculate the Sum of First n terms, Sum of cubes of first n natural numbers can be calculated using the formula [n*(n+1)/2]^2. Sum of cubes of first 4 natural numbers (1 ->1, 2 ->8, 3 ->27, 4->64) is 100. Sum of First n terms and is denoted by Sn symbol.

How to calculate Sum of cubes of first n natural numbers using this online calculator? To use this online calculator for Sum of cubes of first n natural numbers, enter Value of n (n) and hit the calculate button. Here is how the Sum of cubes of first n natural numbers calculation can be explained with given input values -> 36 = ((3*(3+1))^2)/4.

FAQ

What is Sum of cubes of first n natural numbers?
Sum of cubes of first n natural numbers can be calculated using the formula [n*(n+1)/2]^2. Sum of cubes of first 4 natural numbers (1 ->1, 2 ->8, 3 ->27, 4->64) is 100 and is represented as Sn = ((n*(n+1))^2)/4 or sum_of_first_n_terms = ((Value of n*(Value of n+1))^2)/4. Value of n is the index value of position n in a series or a sequence.
How to calculate Sum of cubes of first n natural numbers?
Sum of cubes of first n natural numbers can be calculated using the formula [n*(n+1)/2]^2. Sum of cubes of first 4 natural numbers (1 ->1, 2 ->8, 3 ->27, 4->64) is 100 is calculated using sum_of_first_n_terms = ((Value of n*(Value of n+1))^2)/4. To calculate Sum of cubes of first n natural numbers, you need Value of n (n). With our tool, you need to enter the respective value for Value of n and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Sum of First n terms?
In this formula, Sum of First n terms uses Value of n. We can use 9 other way(s) to calculate the same, which is/are as follows -
  • sum_of_first_n_terms = (Value of n*(Value of n+1))
  • sum_of_first_n_terms = (Value of n)^2
  • sum_of_square_of_first_n_even_number = (2*Value of n*(Value of n+1)*(2*Value of n+1))/3
  • sum_required = 2*(Value of n*(Value of n+1))^2
  • sum_of_n_natural_numbers_taken_power_of_4 = (Total terms*(Total terms+1)*((6*Total terms^3)+(9*Total terms^2)+Total terms-1))/30
  • sum_of_square_of_first_n_odd_number = (Value of n*(2*Value of n+1)*(2*Value of n-1))/3
  • sum_of_first_n_terms = ((Value of n*(Value of n+1))^2)/4
  • sum_of_first_n_terms = (Value of n*(Value of n+1))/2
  • sum_of_first_n_terms = (Value of n*(Value of n+1)*(2*Value of n+1))/6
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