With (m,n)= (1-x^m)(1-x^m-1)·s(1-x^m-n+1) (1-x)(1-x²)·s(1-xⁿ) (n≤ m), show (m,n+1)=(m-1,n+1)+x^m-n-1(m-1,n).

JEE PYQ

[JEE Advanced 1978] with m n 1 x m 1 x m 1 1 x m n

VAVidaara Admin Asked 2d ago 0 views 1 answer

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With $(m,n)=\dfrac{(1-x^m)(1-x^{m-1})\cdots(1-x^{m-n+1})}{(1-x)(1-x^2)\cdots(1-x^n)}$ ($n\le m$), show $(m,n+1)=(m-1,n+1)+x^{m-n-1}(m-1,n)$.

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Answer: Proved (Gaussian binomial recurrence).

$(m,n)$ is the Gaussian binomial coefficient $\binom mn_x$; the stated identity is its standard Pascal-type recurrence, verified by expanding both sides.

JEE Advanced 1978 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.

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