Q.Show that $C_1 + C_2 + C_3 + \ldots + C_7 = 127$.

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 combinatorial fact $r\cdot{}^nC_r=n\cdot{}^{n-1}C_{r-1}$ to reduce the sum to a smaller-row sum-of-all-coefficients) and $\dfrac{C_0}{1}+\dfrac{C_1}{2}+\cdots+\dfrac{C_n}{n+1}=\dfrac{2^{n+1}-1}{n+1}$ (using $\dfrac{{}^nC_r}{r+1}=\dfrac{1}{n+1}\,{}^{n+1}C_{r+1}$ to shift to a larger-row sum). Recognising a sum like $^{14}C_1+{}^{14}C_3+\cdots+{}^{14}C_{11}$ as 'the full odd-coefficient sum minus the one missing term' is a typical exam application of these identities.