Binomial Coefficients

Naming the coefficients. The coefficients $^nC_0,{}^nC_1,{}^nC_2,\ldots,{}^nC_n$ occurring in the expansion of $(a+b)^n$ are called the binomial coefficients, and for brevity are written $C_0,C_1,C_2,\ldots,C_n$.

Sum of all the binomial coefficients equals $2^n$. Start from $(1+x)^n={}^nC_0x^0+{}^nC_1x^1+{}^nC_2x^2+\cdots+{}^nC_nx^n$ ... (i). Substituting $x=1$: $(1+1)^n={}^nC_0+{}^nC_1+\cdots+{}^nC_n$, i.e. $2^n=C_0+C_1+C_2+\cdots+C_n$. So the sum of all the binomial coefficients is $2^n$.

Sum of the even-placed coefficients equals the sum of the odd-placed coefficients, each equal to $2^{n-1}$. Substituting $x=-1$ into (i): $(1-1)^n={}^nC_0-{}^nC_1+{}^nC_2-\cdots+(-1)^n{}^nC_n$, i.e. $0=C_0-C_1+C_2-C_3+\cdots+(-1)^nC_n$, so $C_0+C_2+C_4+\cdots=C_1+C_3+C_5+\cdots$. Call the even-indexed coefficients $C_0,C_2,C_4,\ldots$ and the odd-indexed ones $C_1,C_3,C_5,\ldots$, and let their common value be $k$: $C_0+C_2+C_4+\cdots=C_1+C_3+C_5+\cdots=k$. Adding the two sums gives every coefficient once, so $k+k=C_0+C_1+C_2+\cdots+C_n=2^n$ (from above), i.e. $2k=2^n$, so $k=2^{n-1}$. Hence the sum of the even coefficients equals the sum of the odd coefficients equals $2^{n-1}$.

Solved Example 1. Show that $C_0+C_1+C_2+\cdots+C_{10}=1024$. Solution. Using $C_0+C_1+\cdots+C_n=2^n$ with $n=10$: $C_0+C_1+\cdots+C_{10}=2^{10}=1024$.

Solved Example 2. Show that $C_0+C_2+C_4+\cdots+C_{12}=C_1+C_3+C_5+\cdots+C_{11}=2048$. Solution. With $n=12$: $C_0+C_1+\cdots+C_{12}=2^{12}=4096$ ... (i). By the even/odd-coefficient identity, $C_0+C_2+\cdots+C_{12}=C_1+C_3+\cdots+C_{11}=k$ ... (ii). Adding the two sums in (ii) accounts for every term in (i), so $2k=4096$, $k=2048$. Hence both sums equal $2048$.

Solved Example 3. Prove $C_1+2C_2+3C_3+\cdots+nC_n=n\cdot2^{n-1}$. Solution. For each $r\ge1$, $r\cdot{}^nC_r=r\cdot\dfrac{n!}{r!(n-r)!}=n\cdot\dfrac{(n-1)!}{(r-1)!(n-r)!}=n\cdot{}^{n-1}C_{r-1}$ (a standard combination identity — pulling a factor of $n$ out of $^nC_r$ leaves $^{n-1}C_{r-1}$, since $r$ cancels one factor of $r!$ in the denominator against the top). So L.H.S. $=C_1+2C_2+\cdots+nC_n=n\left[{}^{n-1}C_0+{}^{n-1}C_1+\cdots+{}^{n-1}C_{n-1}\right]=n\left[C_0+C_1+\cdots+C_{n-1}\right]$ (now reading these as the binomial coefficients of the smaller power $n-1$) $=n\cdot2^{n-1}=$ R.H.S., using the sum-of-all-coefficients identity applied to $n-1$ in place of $n$. …