Q.Show that $C_0+2C_1+3C_2+4C_3+\ldots+(n+1)C_n=(n+2)2^{n-1}$.

The coefficients $^nC_0,{}^nC_1,{}^nC_2,\ldots,{}^nC_n$ that appear in the expansion of $(a+b)^n$ are called the binomial coefficients, usually abbreviated $C_0,C_1,\ldots,C_n$ once the value of $n$ is fixed by context. Two identities about these coefficients, obtained just by plugging specific numbers into the expansion of $(1+x)^n$, are used repeatedly throughout this chapter's exercises. Setting $x=1$ turns the expansion into $2^n=C_0+C_1+C_2+\cdots+C_n$, so all the binomial coefficients of a given row sum to $2^n$. Setting $x=-1$ turns it into $0=C_0-C_1+C_2-C_3+\cdots$, which rearranges to show the coefficients in the even positions sum to the same value as the coefficients in the odd positions; adding those two equal sums must reproduce the total $2^n$, so each half is exactly $2^{n-1}$ — the even-placed coefficients sum to $2^{n-1}$, and so, separately, do the odd-placed ones. These two identities are the building blocks for a family of further results proved from them, such as $C_1+2C_2+3C_3+\cdots+nC_n=n\cdot2^{n-1}$ (using the combinator …