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

Explanation

Since, n is an even positive integer, we can wirte
`n =2, , m = 1,2,3….`
Also , ` k = ( 3n)/2 = ( 3(2m))/2 = 3m` therefore `S = overset(3m)underset(r=1)sum(-3)^(r-1).^(6m)C_(2r-1)`
i.e ` S = (-3)^(0).^(6m)C_(3) +... (-3)^(3m-1) . ^(6m)C_(3m-1)`
From the binomial expanision , we write
`(1+x)^(6m)=^(6m)C_(0)+(-3)^(6m)C_(3)+......+(-3)^(3m-1). ^(6m)C_(3m-1)`
` (1-x)^(6m) = ^(6m)C_(0) + ^(6m)C_(1)(-x)+^(6m)C_(2) (-x)^(2)+....+^(6m)C_(6m-1) (-x)^(6m-1)+ ^(6m)C_(6m) (-x)^(6m)`
on subtracting Eq. (iii) from Eq. (ii) , we get
` (1+x)^(6m) - (1-x)^(6m)=2[.^(6m)C_(1)x+ ^(6m)C_(3)x^(3) + ^(6m)C_(5)x^(5x) +...+^(6m) C_(6m-1)x^(6m-1) ]`
` Rightarrow ((1+x)^(6m)-(1-x)^(6m))/(2x)=^(6m)C_(1)+^(6m)C_(3)x^(2)+ ^(6m)C_(5)x^(4) +.....+ ^(6m)C_(6m-1)x^(6m-2)`
Let ` x^(2)=y`
` Rightarrow ((1+sqrty)^(6m)-(1-sqrty)^(6m))/(2sqrty)= ^(6m)C_(1)+^(6m)C_(3)y`
` + ^(6m)C_(5)y^(2)+...+^(6m)C_(6m-1)y^(3m-1)`
For the required sum we have to put y=-3 in RHS.
`S=((1+sqrt3)^(6m)-(1-sqrt(-3))^(6m))/(2sqrt(-3))`
` = ((1+isqrt3)^(6m)-(1-sqrt3)^(6m))/(2isqrt3)`
Let `z= 1 sqrt3= r(cos theta + i sin theta)`
` Rightarrow r= |z|= sqrt(1 +3) =2`
`and theta = pi //3`
Now,` z^(6m)(cos 6m theta) + i sin 6m theta)`
and `(overline(z))^(6m) = r^(6m) (cos 6m theta - i sin 6m theta)`
` Rightarrow z^(6m)- overline(z)^(6m) = r^(6m) (2 isin 6m theta)`
Frow .Eq. (i) ,
` S = (z^(6m)-overlinez^(6m))/(2isqrt3)= (r^(6m)(2isin6m theta))/(2isqrt3)`
`= (2^(6m)sin 6m theta)/(sqrt3)`
`0 "as" m inz, and theta= pi//3`