Q.If $\binom{n}{0} + 2\binom{n}{1} + 3\binom{n}{2} + \cdots + (n+1)\binom{n}{n} = 126$ , then $n$ is

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.If (n0)+2(n1)+3(n2)+⋯+(n+1)(nn)=126\binom{n}{0} + 2\binom{n}{1} + 3\binom{n}{2} + \cdots + (n+1)\binom{n}{n} = 126(0n​)+2(1n​)+3(2n​)+⋯+(n+1)(nn​)=126, then nnn is