# Calculate 16C0 + 16C2 + 16C4 + ... + 16C16?

sciencesolve | Certified Educator

You may also use alternative method to solve the problem.

Notice that you need to find the sum of even binomial coefficients, hence the terms of binomial `(a+b)^n`  needs to be `a=1, b=1`  and `n=16` .

You should consider binomial `(1+1)^16 = C_16^0 + C_16^1 + ... + C_16^16` You should consider the binomial `(1-1)^16 = C_16^0- C_16^1 + ... + C_16^16` You need to add `(1-1)^16 ` to `(1+1)^16`  such that:

`(1-1)^16+ (1+1)^16 = C_16^0 - C_16^1 + ... + C_16^16 + C_16^0 + C_16^1 + ... + C_16^16`

Reducing odd terms yields:

`(1-1)^16 + (1+1)^16 = 2 (C_16^0 + C_16^2 + ... + C_16^16)`

`0^16 + 2^16 = 2*(C_16^0 + C_16^2 + ... + C_16^16)`

You need to divide by 2 both sides such that:

`(C_16^0 + C_16^2 + ... + C_16^16) = 2^16/2`

`(C_16^0 + C_16^2 + ... + C_16^16) = 2^15`

Hence, evaluating the sum of even binomial coefficients yields `(C_16^0 + C_16^2 + ... + C_16^16) = 2^15` .

justaguide | Certified Educator

The value of 16C0+16C2+16C4+...+16C16 has to be calculated.

nCr = `(n!)/(r!*(n - r)!)`

16C0 + 16C2 + 16C4 + ... + 16C16

=> 16C0 + 16C2 + 16C4 + 16C6 + 16C8 + 16C10 + 16C12 + 16C14 + 16C16

=> 1 + 120 + 1820 + 8008 + 12870 + 8008 + 1820 + 120 + 1

=> 32768

The value of 16C0 + 16C2 + 16C4 + ... + 16C16 = 32768