Alg10.15

In the expansion of $(1+a)^n\,$ if the coefficient of $a^{r-1},a^r,a^{r+1}\,$ are in AP.then prove that $n^2-n(4r+1)+4r^2-2=0\,$

If $x,y,z\,$ are in AP,then we know that $2y=x+z\,$

In the expansion of $(1+a)^n\,$ ,the coefficients of $a^{r-1},a^r,a^{r+1}\,$ are $n\!C_{r-1},n\!C_r,n\!C_{r+1}\,$ respectively.

Therefore $2 n\!C_r=n\!C_{r-1}\cdot n\!C_{r+1}\,$

$2=\frac{n\!C_{r-1}}{n\!C_r}+\frac{n\!C_{r+1}}{n\!C_r}\,$ Let this equation be 1.

$\frac{n\!C_{r-1}}{n\!C_r}=\frac{n!}{(r-1)!(n-r+1)!}\cdot\frac{r!\cdot(n-r)!}{n!}=\frac{r}{n-r+1}\,$

Similarly $\frac{n\!C_{r+1}}{n\!C_r}=\frac{n!}{(r+1)!(n-r-1)!}\cdot\frac{r!\cdot (n-r)!}{n!}=\frac{n-r}{r+1}\,$

From the 1, $2=\frac{r}{n-r+1}+\frac{n-r}{r+1}\,$

$2(n-r+1)(r+1)=r(r+1)+(n-r)(n-r+1)\,$

Simplifying $n^2-n(4r+1)+4r^2-2=0\,$

Hence the required.