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

Explanation

`S = underset(r=t)overset(k)sum (-3)^(r-1) .^(3n)C_(2r-1), k = (3n)/(2)` and n is even
Let `k = (3(2m))/(2) = 3m`
Then, `S = underset(r=1)overset(3m)sum(-3)^(r-1) xx .^(6m)C_(2r-1)`
`= .^(6m)C_(1)-3.^(6m)C_(3)+3^(2).^(6m)C_(5)-"……"(-3)^(3m-1).^(6m)C_(6m-1)`
`= 1/(sqrt(3))[sqrt(3).^(6m)C_(1)-(sqrt(3))^(3).^(6m)C_(3) + (sqrt(3))^(5).^(6m)C_(5)-"......"`
` + (-1)^(3m-1)(sqrt(3))^(6m-1).^(6m)C_(6m-1)]`
There is an alternate sign series with odd binomial coefficients.
Hence, we should replace x by `sqrt(3)i` in `(1+x)^(6m)`. Therefore,
`(1+sqrt(3)i)^(6m) = .^(6m)C_(0)+.^(6m)C_(1)(sqrt(3)i)+.^(6m)C_(2)(sqrt(3)i)^(2)+.^(6m)C_(3)(sqrt(3)i)^(3)+"...."+.^(6m)C_(6m)(sqrt(3)i)^(6m)`
`rArr sqrt(3) xx .^(6m)C_(1) -(sqrt(3))^(3).^(6m)C_(3)+(sqrt(3))^(5).^(6m)C_(5)+"...."`
= Imaginary part in `(1+sqrt(3)i)^(6m)`
`= "Im"[2^(6m)(1/2+(sqrt(3))/(2))^(6m)]`
` = "Im"[2^(6m)(cos2mpi + i sin 2m pi)]`
` = "Im" [2^(6m)] = 0`
`rArr S = 0`