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sum(i=1)^20((^20C(i-1))/(^20C+i+^20C(i-1...

`sum_(i=1)^20((^20C_(i-1))/(^20C+_i+^20C_(i-1))=k/21` then find value of k. (a) `400` (b) `100` (c) `50` (d) `200`

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