Prove that`sum_(r=1)^k(-3)^(r-1)C(3n,2r-1)=0`, where `k=(3n)/2` and `n` is an even positive integer.

Explanation

Given , n is an even positive integer .
Let ` n = 2m, therefore k = 3m, n in N `
LHS ` sum_(r=1)^(k) (-3)^(r-1) *""^(3n)C_(2r-1) = sum_(r=1)^(3m) (-3)^(r-1)" " ^(6m)C_(2r-1)`
` = ""^(6m)C_(1)-3*""^(6m)C_(3)+ 3^(2) *""^(6m)C_(5)`
` -...+ (-3)^(3m-16m)C_(6m-1)` ...(i)
Consider `1+i sqrt(3)^(6m)= ""^(6m)C_(0) + ""^(6m)C_(1) (isqrt(3)) + ""^(6m)C_(2) (isqrt(3))^(2)`
` + ""^(6m)C_(3) (isqrt(3))^(3) + ""^(6m)C_(4) (isqrt(3))^(4)+""^(6m)C_(5) (isqrt(3))^(5)`
` + ...+ ""^(6m)C_(6m-1) (isqrt(3))^(6m-1) + ""^(6m)C_(6m) (isqrt(3))^(6m) `...(ii)
Now , `(1 + i sqrt(3))^(6m) = {(-2)((-1-isqrt(3))/((2)))}^(6m) = (-2omega^(2) )^(6m)`
` = 2^(6m) ," where " omega` is cube root of unity.
Then,Eq .(ii) canbe written as
`2^(6m) = { ""^(6m)C_(0)- ""^(6m)C_(2) *3 + ""^(6m)C_(4) *3^(2) `
` - ...+ (-3)^(3m) *""^(6m)C_(6m)} + isqrt(3) {""^(6m)C_(1) - ""^(6m)C_(3) *3 + `
`+""^(6m)C_(5) *3^(2) -...+(-3)^(3m-1)*""^(6m)C_(6m-1)}`
On comparing the imaginary part on both sides , we get
`sqrt(3)(""^(6m)C_(1) -3+ ""^(6m)C_(3) + 3^(2) * ""^(6m)C_(5) -...+ (-3)^(3m-1)*""^(6m)C_(6m-1)) = 0 `
or `""^(6m)C_(1) -3+ ""^(6m)C_(3) + 3^(2) * ""^(6m)C_(5) -...+ (-3)^(3m-1)*""^(6m)C_(6m-1) = 0 `
`rArr sum_(r=1)^(3m) (-3)^(3m-1)*""^(3n)C_(2r-1) = 0 `
or ` sum_(r=1)^(k) (-3)^(r-1) *""^(3n)C_(2r-1) = 0` , where n = 2m nad k = 3m .