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

`sum_(i=1)^20 [(.^20C_(i-1))/(.^20C_i+.^20C_(i-1))]^3=k/21` then find the value of k. (a) 400 (b) 100 (c) 50 (d) 200

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