MathematicsEasy64×since 2002Q3002∑\matrixi,j=0\cri≠j\crnnCi nCj\sum\limits_{\matrix{ {i,j = 0} \cr {i \ne j} \cr } }^n {{}^n{C_i}\,{}^n{C_j}}\matrixi,j=0\cri=j\cr∑nnCinCj is equal toA22n−2nCn2^{2 n}-{ }^{2 n} C_{n}22n−2nCnB22n−1−2n−1Cn−1{2^{2n - 1}} - {}^{2n - 1}{C_{n - 1}}22n−1−2n−1Cn−1C22n−122nCn2^{2 n}-\frac{1}{2}{ }^{2 n} C_{n}22n−212nCnD22n−1+2n−1Cn{2^{2n - 1}} + {}^{2n - 1}{C_n}22n−1+2n−1CnCheck answerSkip