Mathematics
Nietzer897
16

the coefficient of x^k*y^n-k; true or false

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(1) Answers
Mckinley731

The coefficient of x^k*y^n-k; true or false=Answer Proof: (x + y)^2 = x^2 + 2xy + y^2. If n is 2, k must by 1; nCk = 2/1 = 2, the coefficient of the xy term.  (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3. If n is 3, k = 1 leaves nCk = 3/1 = 3, the coefficient of the xy^2 term. k = 2 fails, as nCk = 3/2, which is NOT the coefficient of the x^2y term.  (x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + x^4. If n is 4, k =1 leaves nCk = 4/1 = 4, the coefficient of the xy^3 term. k = 2 and k = 3 both fail (see (x + y)^3 expansion above).  (x + y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5. If n is 5, k = 1 leaves nCk = 5/1 = 5, matching the coefficient of the xy^4 term. k = 2, k = 3, and k = 4 all fail.  In summary, your question as a whole is FALSE for all real numbers k.  The only value of k that makes your statement true is k = 1, which means the expansion of (x + y)^n will have a coefficient of the x^ky^(n - k) term matching nCk only for that value of k alone. And that's just considering whole number exponents.  (EDIT: I forgot one other possibility: n = k can work: e.g., if k = 2 in the expansion of (x + y)^2, nCk = 2/2 = 1, which matches the coefficient of the x^2 term. The same applies for other expansions of (x + y)^n where n is a whole number 3 or higher; as long as n = k, the x^k term has the same coefficient as the quotient of nCk.)

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