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This question came about after reading this other question here.

Show that $$\sum_{r=1}^n a^{r-1}\left[\binom n{r-1}-(a+1)^{n-r}\right]=0$$ without expanding the summation in full.

If we expand the summation in full we have

$$\begin{align} \sum_{r=1}^n \binom n{r-1}a^{r-1} &=\sum_{r=0}^{n-1}\binom nr a^r\\ &=\sum_{r=0}^n \binom nr a^r-a^n\\ &=(a+1)^n-a^n\end{align}$$ and $$\begin{align} \sum_{r=1}^n (a+1)^{n-r}a^{r-1} &=\frac {(a+1)^n}a\sum_{r=1}^n\left(\frac a{a+1}\right)^r\\ &=\frac {(a+1)^n}a\cdot \frac a{a+1}\cdot\frac {1-\left(\frac a{a+1}\right)^n}{1-\frac a{a+1}}\\ &=(a+1)^n-a^n\end{align}$$ which are equal to each other.

Hypergeometricx
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1 Answers1

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Having given it further thought, here's one approach:

$$\begin{align} \sum_{r=1}^n a^{r-1}(a+1)^{n-r} &=\sum_{r=1}^na^{r-1}\sum_{j=0}^{n-r}\binom {n-r}ja^j\\ &=\sum_{r=1}^n\sum_{j=0}^{n-r}\binom {n-r}ja^{r+j-1}\\ &=\sum_{r=1}^n\sum_{k=r-1}^{n-1}\binom {n-r}{k+1-r}a^k &&(k=r+j-1)\\ &=\sum_{r=1}^n\sum_{k=r}^n\binom {n-r}{k-r}a^{k-1}\\ &=\sum_{k=1}^n a^{k-1}\sum_{r=1}^k\binom {n-r}{k-r}\\ &=\sum_{k=1}^n a^{k-1}\sum_{r=1}^k\binom {n-r}{n-k}\\ &=\sum_{k=1}^n a^{k-1}\sum_{s=n-k}^{n-1}\binom s{n-k} &&(s=n-r)\\ &=\sum_{k=1}^n a^{k-1}\binom n{n-k+1}\\ &=\sum_{k=1}^n a^{k-1}\binom n{k-1}\\ &=\sum_{r=1}^n a^{r-1}\binom n{r-1}\\ \sum_{r=1}^n a^{r-1}\left[\binom n{r-1}-(a+1)^{n-r}\right]&=0\end{align}$$

Hypergeometricx
  • 22,657