Question

# The value of (300)(3010)−(301)(3011)+(302)(3012)+⋯+(3020)(3020)

Solution

## The correct option is D (1−x)30=30C0x0−30C1x1+30C2x2+⋯(−1)3030C30x30 .............(i) (1+x)30=30C0x30+30C1x29+30C2x28+⋯30C10x0+⋯+30C30x0 .............. (ii)  Multiplying (i) and (ii) and equating the coefficient of x20 on both sides, we get required sum = coefficient of x20 in (1−x2)30=30C10

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