  Question

(30C0)(30C10)−(30C1)(30C11)+.....(30C20)(30C30) is equal to

A   B   C   D   Solution

The correct option is A Each of the terms has two coefficients. We can re-write the terms as follows (30C0)(30C20)−30C1(30C19)....+30C2030C0 If we look at the terms now, the sum of bottom numbers in the coefficients add up to 20. (e.g.30C030C20; here 0 + 20 = 20) At this point we should be able to guess that each term is formed by multiplying two terms from two binomial expansions. One of them should have negative terms also. Let us consider the expansion of (1+x)30(1−x)30, (1+x)30(1−x)30 =(30C0+30C1x+30C2x2.....30C10x10+.....30C30x30)×(30C0−30C1x+30C2x2.....30C10x10+.....30C30x30) We will find the coefficient of x20 = 30C030C20+30C1×(−30C19)+30C230C18.....(30C20)(30C0) This is the required sum. We can say that the required sum is the coefficient x20 in the expansion of (1+x)30(1−x)30 or (1−x2)30. Coefficient of x20 in the expansion of (1−x2)30=30C10 [Tr+1=30Cr×(−x2)r ⇒ 2r = 20 ⇒ r = 10 ⇒ coefficient = 30C10 Mathematics

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