Binomial theorem equation

Question

I read in this answer the following equation:

$\displaystyle \sum^n_{k=1} \binom {n-1}{k-1}p^{k-1}(1-p)^{n-k} = (p + (1-p))^{n-1} = 1$

I was trying to calculate the left side my self but my result is different:

$$ \sum^n_{k=1} \binom {n-1}{k-1}p^{k-1}(1-p)^{n-k} = p^{n-1} + \sum^{n-1}_{k=1} \binom {n-1}{k-1}p^{k-1}(1-p)^{n-k} = p^{n-1} + \sum^{n-1}_{k=1} \binom {n-1}{k-1}p^{k-1}(1-p)^{n-1-(k-1)} = $$ $$ m= k-1 $$ $$ p^{n-1} + \sum^{n-1}_{m=0} \binom {n-1}{m}p^{m}(1-p)^{n-1-m} = $$

From binomial theorem: $\displaystyle (x+y)^{n}=\sum^{n}_{k=0} \binom {n}{k}x^{k}y^{n-k}$

$$ p^{n-1} + \sum^{n-1}_{m=0} \binom {n-1}{m}p^{m}(1-p)^{n-1-m} = p^{n-1} +1 $$

Where is my mistake?

If $m= k-1$, then $m$ iterates from $0$ to $n-2$, so the sum should be $\sum^{n-2}_{m=0}$. — The Phenotype, May 25 '18 at 15:04
Well, if $k=1,2,3,\ldots, n-1$ and $m$ is $k$ translated by $1$, then $m=1-1,2-1,3-1,\ldots, n-1-1=0,1,2,\ldots, n-2$. — The Phenotype, May 25 '18 at 15:07

score 1 · Accepted Answer · edited May 25 '18 at 15:13

1

After variable substitution $$ m= k-1 $$ you must decrease both summation bounds: $$p^{n-1} + \sum^{n-2}_{m=0}\ldots$$

edited May 25 '18 at 15:13

The Phenotype

5,300

answered May 25 '18 at 15:08

Vladislav Alexeev

69

Binomial theorem equation

1 Answers1