Sum of squares of binomial coefficients
Last Updated :
28 Mar, 2023
Given a positive integer n. The task is to find the sum of square of Binomial Coefficient i.e
nC02 + nC12 + nC22 + nC32 + ……… + nCn-22 + nCn-12 + nCn2
Examples:
Input : n = 4
Output : 70
Input : n = 5
Output : 252
Method 1: (Brute Force)
The idea is to generate all the terms of binomial coefficient and find the sum of square of each binomial coefficient.
Below is the implementation of this approach:
C++
#include<bits/stdc++.h>
using namespace std;
int sumofsquare(int n)
{
int C[n+1][n+1];
int i, j;
for (i = 0; i <= n; i++)
{
for (j = 0; j <= min(i, n); j++)
{
if (j == 0 || j == i)
C[i][j] = 1;
else
C[i][j] = C[i-1][j-1] + C[i-1][j];
}
}
int sum = 0;
for (int i = 0; i <= n; i++)
sum += (C[n][i] * C[n][i]);
return sum;
}
int main()
{
int n = 4;
cout << sumofsquare(n) << endl;
return 0;
}
|
Java
import static java.lang.Math.*;
class GFG{
static int sumofsquare(int n)
{
int[][] C = new int [n+1][n+1] ;
int i, j;
for (i = 0; i <= n; i++)
{
for (j = 0; j <= min(i, n); j++)
{
if (j == 0 || j == i)
C[i][j] = 1;
else
C[i][j] = C[i-1][j-1]
+ C[i-1][j];
}
}
int sum = 0;
for (i = 0; i <= n; i++)
sum += (C[n][i] * C[n][i]);
return sum;
}
public static void main(String[] args)
{
int n = 4;
System.out.println(sumofsquare(n));
}
}
|
Python3
def sumofsquare(n) :
C = [[0 for i in range(n + 1)]
for j in range(n + 1)]
for i in range(0, n + 1) :
for j in range(0, min(i, n) + 1) :
if (j == 0 or j == i) :
C[i][j] = 1
else :
C[i][j] = (C[i - 1][j - 1] +
C[i - 1][j])
sum = 0
for i in range(0, n + 1) :
sum = sum + (C[n][i] *
C[n][i])
return sum
n = 4
print (sumofsquare(n), end="\n")
|
C#
using System;
class GFG {
static int sumofsquare(int n)
{
int[,] C = new int [n+1,n+1] ;
int i, j;
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.Min(i, n); j++)
{
if (j == 0 || j == i)
C[i,j] = 1;
else
C[i,j] = C[i-1,j-1]
+ C[i-1,j];
}
}
int sum = 0;
for (i = 0; i <= n; i++)
sum += (C[n,i] * C[n,i]);
return sum;
}
public static void Main()
{
int n = 4;
Console.WriteLine(sumofsquare(n));
}
}
|
PHP
<?php
function sumofsquare($n)
{
$i; $j;
for ($i = 0; $i <= $n; $i++)
{
for ($j = 0; $j <= min($i, $n); $j++)
{
if ($j == 0 || $j == $i)
$C[$i][$j] = 1;
else
$C[$i][$j] = $C[$i-1][$j-1]
+ $C[$i-1][$j];
}
}
$sum = 0;
for ($i = 0; $i <= $n; $i++)
$sum += ($C[$n][$i] * $C[$n][$i]);
return $sum;
}
$n = 4;
echo sumofsquare($n), "\n";
?>
|
Javascript
<script>
function sumofsquare(n)
{
let C = new Array(n+1);
for (let i = 0; i < C.length; i++) {
C[i] = new Array(2);
}
let i, j;
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.min(i, n); j++)
{
if (j == 0 || j == i)
C[i][j] = 1;
else
C[i][j] = C[i-1][j-1]
+ C[i-1][j];
}
}
let sum = 0;
for (i = 0; i <= n; i++)
sum += (C[n][i] * C[n][i]);
return sum;
}
let n = 4;
document.write(sumofsquare(n));
</script>
|
Output:
70
Time Complexity: O(n2)
Space Complexity: O(n2)
Method 2: (Using Formula)
= 
= 
Proof,
We know,
(1 + x)n = nC0 + nC1 x + nC2 x2 + ......... + nCn-1 xn-1 + nCn-1 xn
Also,
(x + 1)n = nC0 xn + nC1 xn-1 + nC2 xn-2 + ......... + nCn-1 x + nCn
Multiplying above two equations,
(1 + x)2n = [nC0 + nC1 x + nC2 x2 + ......... + nCn-1 xn-1 + nCn-1 xn] X
[nC0 xn + nC1 xn-1 + nC2 xn-2 + ......... + nCn-1 x + nCn]
Equating coefficients of xn on both sides, we get
2nCn = nC02 + nC12 + nC22 + nC32 + ......... + nCn-22 + nCn-12 + nCn2
Hence, sum of the squares of coefficients = 2nCn = (2n)!/(n!)2.
Also, (2n)!/(n!)2 = (2n * (2n – 1) * (2n – 2) * ….. * (n+1))/(n * (n – 1) * (n – 2) *….. * 1).
Below is the implementation of this approach:
C++
#include<bits/stdc++.h>
using namespace std;
int factorial(int start, int end)
{
int res = 1;
for (int i = start; i <= end; i++)
res *= i;
return res;
}
int sumofsquare(int n)
{
return factorial(n+1, 2*n)/factorial(1, n);
}
int main()
{
int n = 4;
cout << sumofsquare(n) << endl;
return 0;
}
|
Java
class GFG{
static int factorial(int start, int end)
{
int res = 1;
for (int i = start; i <= end; i++)
res *= i;
return res;
}
static int sumofsquare(int n)
{
return factorial(n+1, 2*n)/factorial(1, n);
}
public static void main(String[] args)
{
int n = 4;
System.out.println(sumofsquare(n));
}
}
|
Python
def factorial(start, end):
res = 1
for i in range(start, end + 1):
res *= i
return res
def sumofsquare(n):
return int(factorial(n + 1, 2 * n)
/factorial(1, n))
n = 4
print(sumofsquare(n))
|
C#
using System;
class GFG {
static int factorial(int start, int end)
{
int res = 1;
for (int i = start; i <= end; i++)
res *= i;
return res;
}
static int sumofsquare(int n)
{
return factorial(n+1, 2*n)/factorial(1, n);
}
public static void Main()
{
int n = 4;
Console.WriteLine(sumofsquare(n));
}
}
|
PHP
<?php
function factorial($start, $end)
{
$res = 1;
for ($i = $start;
$i <= $end; $i++)
$res *= $i;
return $res;
}
function sumofsquare($n)
{
return factorial($n + 1,
2 * $n) /
factorial(1, $n);
}
$n = 4;
echo sumofsquare($n), "\n";
?>
|
Javascript
<script>
function factorial(start, end)
{
let res = 1;
for (let i = start; i <= end; i++)
res *= i;
return res;
}
function sumofsquare(n)
{
return parseInt
(factorial(n+1, 2*n)/factorial(1, n), 10);
}
let n = 4;
document.write(sumofsquare(n));
</script>
|
Output:
70
Time Complexity: O(n)
Auxiliary Space: O(1)
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