Given a positive ineteger n. The task is to find the sum of product of consecutive binomial coefficient i.e
nC0*nC1 + nC1*nC2 + ….. + nCn-1*nCn
Input : n = 3 Output : 15 3C0*3C1 + 3C1*3C2 +3C2*3C3 = 1*3 + 3*3 + 3*1 = 3 + 9 + 3 = 15 Input : n = 4 Output : 56
Method 1: The idea is to find all the binomial coefficients up to nth term and find the sum of the product of consecutive coefficients.
Below is the implementation of this approach:
(1 + x)n = nC0 + nC1*x + nC2*x2 + …. + nCn*xn … (1)
(1 + 1/x)n = nC0 + nC1/x + nC2/x2 + …. + nCn/xn … (2)
Multiplying (1) and (2), we get
(1 + x)2n/xn = (nC0 + nC1*x + nC2*x2 + …. + nCn*xn) * (nC0 + nC1/x + nC2/x2 + …. + nCn/xn)
(2nC0 + 2nC1*x + 2nC2*x2 + …. + 2nCn*xn)/xn = (nC0 + nC1*x + nC2*x2 + …. + nCn*xn) * (nC0 + nC1/x + nC2/x2 + …. + nCn/xn)
Now, find the coefficient of x in LHS,
Observe rth term of expansion in numerator is 2nCrxr.
To find the coefficient of x in (1 + x)2n/xn, r should be n + 1, because power of x in denominator will reduce it.
So, coefficient of x in LHS = 2nCn + 1 or 2nCn – 1
Now, find the coefficient of x in RHS,
r th term of first expansion of multiplication is nCr * xr
t th term of second expansion of multiplication is nCt / xt
So term after multiply will be nCr * xr * nCt / xt or
nCr * nCt * xr / xt
Put r = t + 1, we get,
nCt+1 * nCt * x
Observe there will be n such term in the expansion of multiply, so t range from 0 to n – 1.
Therefor, coefficient of x in RHS = nC0*nC1 + nC1*nC2 + ….. + nCn-1*nCn
Comparing coefficient of x in LHS and RHS, we can say,
nC0*nC1 + nC1*nC2 + ….. + nCn-1*nCn = 2nCn – 1
Below is implementation of this approach:
# Python3 Program to find sum of product
# of consecutive Binomial Coefficient.
MAX = 100;
# Find the binomial coefficient
# up to nth term
def binomialCoeff(n, k):
C =  * (k + 1);
C = 1; # nC0 is 1
for i in range(1, n + 1):
# Compute next row of pascal triangle
# using the previous row
for j in range(min(i, k), 0, -1):
C[j] = C[j] + C[j – 1];
# Return the sum of the product of
# consecutive binomial coefficient.
return binomialCoeff(2 * n, n – 1);
# Driver Code
n = 3;
# This code is contributed by mits
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