Python Program for Sum of squares of first n natural numbers
Given a positive integer N. The task is to find 12 + 22 + 32 + ….. + N2.
Examples:
Input : N = 4
Output : 30
12 + 22 + 32 + 42
= 1 + 4 + 9 + 16
= 30
Input : N = 5
Output : 55
Method 1: O(N) The idea is to run a loop from 1 to n and for each i, 1 <= i <= n, find i2 to sum.
python3
def squaresum(n):
sm = 0
for i in range ( 1 , n + 1 ):
sm = sm + (i * i)
return sm
n = 4
print (squaresum(n))
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Method 2: O(1)
Proof:
We know,
(k + 1)3 = k3 + 3 * k2 + 3 * k + 1
We can write the above identity for k from 1 to n:
23 = 13 + 3 * 12 + 3 * 1 + 1 ......... (1)
33 = 23 + 3 * 22 + 3 * 2 + 1 ......... (2)
43 = 33 + 3 * 32 + 3 * 3 + 1 ......... (3)
53 = 43 + 3 * 42 + 3 * 4 + 1 ......... (4)
...
n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ......... (n - 1)
(n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ......... (n)
Putting equation (n - 1) in equation n,
(n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1
= (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1
By putting all equation, we get
(n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/6 = Σ k2
python3
def squaresum(n):
return (n * (n + 1 ) * ( 2 * n + 1 )) / / 6
n = 4
print (squaresum(n))
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Time complexity: O(1)
Auxiliary space: O(1)
Avoiding early overflow: For large n, the value of (n * (n + 1) * (2 * n + 1)) would overflow. We can avoid overflow up to some extent using the fact that n*(n+1) must be divisible by 2.
Python3
def squaresum(n):
return (n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3
n = 4
print (squaresum(n))
|
Time complexity: O(1) because constant operations are being performed
Auxiliary space: O(1)
Approach: list comprehension
- Take input from the user as a positive integer N using the input() function.
- Convert the input string to an integer using the int() function and store it in a variable N.
- Use a list comprehension to create a list of squares of numbers from 1 to N. The list comprehension should look like this: [i*i for i in range(1, N+1)]. This creates a list of squares of numbers from 1 to N.
- Use the sum() function to find the sum of all elements in the list. Store the result in a variable sum_of_squares.
- Print the result using the print() function.
Python3
N = 5
squares_list = [i * i for i in range ( 1 , N + 1 )]
sum_of_squares = sum (squares_list)
print ( "Sum of squares of first" , N, "natural numbers is" , sum_of_squares)
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Output
Sum of squares of first 5 natural numbers is 55
The time complexity using a list comprehension is O(n)
The auxiliary space or space complexity is also O(n)
Approach: iterative
We can use a loop to iterate through the first n natural numbers and calculate the sum of their squares.
Steps:
- Initialize a variable sum to 0.
- Use a loop to iterate through the first n natural numbers, i.e., from 1 to n.
- Within the loop, calculate the square of the current number and add it to the sum.
- After the loop completes, print the value of sum.
Python3
n = 4
sum = 0
for i in range ( 1 , n + 1 ):
sum + = i * * 2
print ( "The sum of squares of first" , n, "natural numbers is" , sum )
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Output
The sum of squares of first 4 natural numbers is 30
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Sum of squares of first n natural numbers for more details!
METHOD 5:Using recursion
APPROACH:
This program calculates the sum of squares of first n natural numbers recursively.
ALGORITHM:
1.Define a function sum_of_squares(n) which takes an integer n as input.
2.Check if n is 1, return 1.
3.Else, calculate the sum of squares recursively by adding n*n with the sum_of_squares of n-1.
4.Set the value of N as 4.
5.Call sum_of_squares function with N as input and store the result in sum_of_squares variable.
6.Print the result.
Python3
def sum_of_squares(n):
if n = = 1 :
return 1
else :
return n * n + sum_of_squares(n - 1 )
N = 4
sum_of_squares = sum_of_squares(N)
print ( "Sum of squares of first" , N, "natural numbers:" , sum_of_squares)
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Output
Sum of squares of first 4 natural numbers: 30
Time Complexity: O(n)
Space Complexity: O(n)
Last Updated :
25 Apr, 2023
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