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Question

# The value of (300)(3010)−(301)(3011)+(302)(3012)...+(3020)(3030) is; where (nr)=nCr

A
(3010)
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B
(3015)
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C
(6030)
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D
(3110)
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Solution

## The correct option is A (3010)To find 30C030C10−30C301C11+30C302C12−...+30C20 30C30 We know that (1+x)30=30C0+30C1x+30C2x2+.....+30C20x20+...30C30x30 .....(1)(x−1)30=30C0x30−30C1x29+....+30C10x20−30C11x19+30C12x18+...30C30x0 .....(2) Multiplying equation (1) and (2),we get (x2−1)30=()×() Equating the coefficients of x20 on both sides, we get 30C10=30C300C10−30C301C11+30C302C12−...+30C20 30C30 ∴ Req.value is 30C10

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