## Thursday, December 6, 2018

### AP Sequence and Binomial Expansion

$\def\D{\displaystyle}$
Let $\D a_0,a_1,a_2,\ldots,a_n$ be the  coefficients of the expansion of $\D (1+x)^n$, and define $\D u_k=\frac{a_{k-1}}{a_{k-1}+a_{k}}, k=1,2,\ldots,n.$
Show that $\D u_{k+1}-u_k= \frac{1}{n+1},$ for $\D k=1,2,\ldots, n-1.$
Hence deduce that $\D u_k+u_{k+2}=2u_{k+1}$ for $\D k=1,2,\ldots,n-2.$

### , Proof:

Since $\D a_k=^nC_k,$
\begin{eqnarray}
a_k&=& \frac{n(n-1)\cdots (n-k+1)}{1\times2\times \cdots \times k}\\
a_{k+1}&=&\frac{n(n-1)\cdots (n-k+1)(n-k)}{1\times2\times \cdots \times k\times (k+1)}.
\end{eqnarray}
$\D (2)\div (1):\frac{a_{k+1}}{a_k}=\frac{n-k}{k+1},$ for $\D k=0,1,\ldots,(n-1).$
Moreover $\D \frac{1}{u_k}= \frac{a_k+a_{k+1}}{a_k}$$\D =1+\frac{a_{k+1}}{a_k}$$\D =1+\frac{n-k}{k+1}$$\D =\frac{k+1+n-k}{k+1} =\frac{n+1}{k+1},$ and hence $\D u_k=\frac{k+1}{n+1}.$
Now $\D u_{k+1}-u_k=\frac{(k+1)+1}{n+1}-\frac{k+1}{n+1}=\frac{1}{n+1}.$
Thus $\D u_1,u_2,\ldots,u_n$ is AP with common difference $\D \frac{1}{n+1}.$
Therefore $\D u_{k+2}-u_{k+1}=u_{k+1}-u_k$ implies
$\D u_k+u_{k+2}=2u_{k+1}.$