JEE Advance - Mathematics (1993 - No. 10)
Prove that $$\sum\limits_{r = 1}^k {{{\left( { - 3} \right)}^{r - 1}}\,\,{}^{3n}{C_{2r - 1}} = 0,} $$ where $$k = \left( {3n} \right)/2$$ and $$n$$ is an even positive integer.
The statement is true due to symmetry properties of binomial coefficients and alternating signs when n is even.
The statement is false because the summation will always result in a non-zero value for even n.
The statement is conditionally true, depending on the divisibility of n by other prime numbers.
The statement can only be proven using advanced complex analysis and contour integration techniques.
The statement requires further clarification on the range of 'r' and its relationship to 'n'.