Q.

If 1×2020×2×2019×3×2018+...........2020×1=2020×2021×k , then the value of k100 is equal to

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answer is 3.37.

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Detailed Solution

Tr=r(2021−r)S=∑r=12020 Tr=2021∑r=12020 r−∑r=12020 r2=2021×2020×20212−2020×2021×40416=2020×202122021−40413=2020×202126063−40413=2020×2021×20226=2020×2021×337It is given that S = 2020 x 2021 x kTherefore, k=337⇒k100=337100=3.37
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If 1×2020×2×2019×3×2018+...........2020×1=2020×2021×k , then the value of k100 is equal to