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

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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 | College Teacher | (Level 2) Distinguished Educator

Posted on

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