Question

# Prove that $$1 \, - \, 2n \, + \, \dfrac{2n(2n \, - \, 1)}{2!} \, - \, \dfrac{2n(2n \, - \, 1)(2n \, - \, 1)}{3!} \, + \, ... \, + \, (-1)^{n \, - \, 1} \, \dfrac{2n(n \, - \, 1)...(n \, + \, 2)}{(n \, - \, 1)}$$$$= \, (-1)^{n \, + \, 1} \, \dfrac{(2n)}{2(n!)^2},$$ where n is a + ive integer.

Solution

## L.H.S.$$= \, S \,= \,1 \, - \, ^{2n}C_1 \, + \, ^{2n}C_2 \, - \,......+ \, (-1)^{n \,-\,1} \, \, ^{2n}C_{n \, - \, 1}$$ n terms Now we know that $$^{2n}C_0 \, = \, 1 \, = \, ^{2n}C_{2n}$$$$^{2n}C_1 \, = \, ^{2n}C_{2n \, - \, 1}$$ etc.$$\therefore \, 2S \, = \, \left ( ^{2n}C_0 \, + \, ^{2n}C_{2n}\right ) \, - \, \left ( ^{2n}C_1 \, + \, ^{2n}C_{2n \, - \, 1} \right ) \, + \, .....+ \,(-1)^{n \, - \, 1} \, \left ( ^{2n}C_{n \, - \, 1} \, + \, ^{2n}C_{n \, + \, 1} \right )$$ $$2S \, =\, ^{2n}C_0 \, - \, ^{2n}C_1 \, + \, ^{2n}C_2 \, - \, ......- \, ^{2n}C_{2n \, - \, 1} \, + \, ^{2n}C_{2n}$$      2n terms Add $$(-1)^n$$   $$^{2n}C_n$$ to both sides to make $$(2n \, + \, 1)$$ terms $$2S \, + \, (-1)^n \, \, ^{2n}C_n \, = C_0 \, - \, C_1 \, + \, C_2 \, - \, + \, C_{2n} \, = \, 0$$[ on putting x = -1 in the expansion of $$(1 + x)^{2n}$$ which contains (2n + 1) terms ] $$\therefore \, S = \, - \, \dfrac{1}{2} \, (-1)^n \, \, \,^{2n}C_n \, = \, (-1)^{n \, + \, 1} \, \, \, \dfrac{(2n)!}{2(n!)^2} \,= \, R.H.S.$$ Mathematics

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