Question

Prove that ` sum_(k=0)^(n) (-1)^(k).""^(3n)C_(k) = (-1)^(n). ""^(3n-1)C_(n)`

Answer 1

We have,
`S = underset(k = 0)overset(n)sum(-1)^(k) . .^(3n)C_(k) = .^(3n)C_(0) - .^(3n)C_(1) + .^(3n)C_(2) + "……" + (-1)^(n)..^(3n)C_(n)`
But
`.^(3n)C_(0) = .^(3n-1)C_(0)`
`-.^(3n)C_(1) = -.^(3n-1)C_(0) - .^(3n-1)C_(1)`
`.^(3n)C_(2) = .^(3n-1)C_(1) .^(3n-1)C_(2)`
`-.^(3n)C_(3) = -.^(3n-1)C_(2) - .^(3n-1)C_(3)`
`{:("......"),("......"):}" "{:("......"),("......"):}`
`(-1)^(n)..^(3n)C_(n) = (-1)^(n).^(3n-1)C_(n-1)+(-1)^(n).^(3n-1)C_(n)`
On adding, we get
`S = (-1)^(n).^(3n-1)C_(n)`.