JEE Main 202126 Feb, Morning shiftMathematicsHardBinomial Theorem and Its Simple Applications131× since 2002Let m, n∈\in∈N and gcd (2, n) = 1. If 30(300)+29(301)+......+2(3028)+1(3029)=n.2m30\left( {\matrix{ {30} \cr 0 \cr } } \right) + 29\left( {\matrix{ {30} \cr 1 \cr } } \right) + ...... + 2\left( {\matrix{ {30} \cr {28} \cr } } \right) + 1\left( {\matrix{ {30} \cr {29} \cr } } \right) = n{.2^m}30(300)+29(301)+......+2(3028)+1(3029)=n.2m, then n + m is equal to __________. (Here (nk)=nCk\left( {\matrix{ n \cr k \cr } } \right) = {}^n{C_k}(nk)=nCk)Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution