JEE Main 202312 Apr, Morning shiftMathematicsMediumBinomial Theorem and Its Simple Applications131× since 2002If 1n+1nCn+1nnCn−1+…+12nC1+nC0=102310\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}n+11nCn+n1nCn−1+…+21nC1+nC0=101023 then nnn is equal to :AOption A. 9BOption B. 6COption C. 7DOption D. 8CheckSkip · show solution