C-13-Q049mediumsingle_mcq1+2+3+⋯+99=?1 + 2 + 3 + \cdots + 99 = ?1+2+3+⋯+99=?a495049504950b485148514851c505050505050d990099009900ব্যাখ্যাUsing n(n+1)2\dfrac{n(n+1)}{2}2n(n+1) with n=99n=99n=99 gives 99⋅1002=4950\dfrac{99\cdot 100}{2}=4950299⋅100=4950.