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