1

$\sum_{r=0}^n C(n-1,r-1)$ = C(n-1,-1) + C(n-1,0) + C(n-1,1)+....+ C(n-1,n-1)

How is the 1st term valid i.e C(n-1,-1). If the 1st term is invalid, then $\sum_{r=0}^n C(n-1,r-1)$ is also invalid. Isn't it? Please explain.

But I know if I take C(n,r) = C(n-1,r) + C(n-1,r-1)

S0, C(n-1,r-1) = C(n,r) - C(n-1,r)

Hence $\sum_{r=0}^n C(n-1,r-1)$ = $\sum_{r=0}^n C(n,r)$ - $\sum_{r=0}^n C(n-1,r)$, which equals to $2^{n-1}$. How is this possible?

RobPratt
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Rajesh Marndi
  • 469

2 Answers2

2

For integral (even complex) $n$ and integral $r$ the following definition holds:

\begin{align*} \binom{n}{r}= \begin{cases} \frac{n(n-1)\cdots (n-r+1)}{r!}&\quad r\geq 0\\ 0&\quad r<0\tag{1} \end{cases} \end{align*}

See for instance formula (5.1) in chapter Binomial Coefficients of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.

From (1) we conclude $\binom{n-1}{-1}=0$.

We obtain \begin{align*} \color{blue}{\sum_{r=0}^n\binom{n-1}{r-1}}&=\sum_{r=1}^n\binom{n-1}{r-1}\tag{2}\\ &=\sum_{r=0}^{n-1}\binom{n-1}{r}\tag{3}\\ &\,\,\color{blue}{=2^{n-1}}\tag{4} \end{align*}

Comment:

  • In (2) we start with index $r=1$, since $\binom{n-1}{-1}=0$ according to (1).

  • In (3) we shift the index $r$ by one to start with $r=0$.

  • In (4) we apply the binomial theorem.

Markus Scheuer
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1

Recall that $$C(A, B)=\frac{A!}{B!(A-B)!}$$

What is often unconsciously over-applied is the recursive factorial definition $$(k-1)!=\frac{k!}{k}$$ Indeed, this is true wherever $k\in \Bbb N$, and the formula you state $$C(n,r)=C(n-1,r)+C(n-1,r-1)$$ flagrantly uses this very definition. However, when we apply it to $k=0$, we arrive at the massive problem that $(-1)!=\frac 10$. The formula still works under that condition, indeed, if we set $r=0$, we have: $$\frac{n!}{0!n!}=\frac{(n-1)!}{0!(n-1)!}+\frac{(n-1)!}{(-1)!n!}\implies\frac{1}{(-1)!n!}=0$$ but clearly the use of an undefined number makes this impossible to use in our sum.

Rhys Hughes
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