If the coefficient of $a^{r - 1}, a^{r}$ and $a^{r + 1}$ in the expansion of $(1 + a)^{n}$ are in arithmetic progression, prove that $n^{2} - n (4 r + 1) + 4 r^{2} - 2 = 0$ .

Open in App

Solution

coefficient of $a r^{- 1}$ in $(1 + a)^{n}$
$^{n} C_{r - 1} (1)^{n - r + 1} (a)^{r - 1}$
$a^{r} =^{n} C_{r} (1)^{n - r} (a)^{r}$

Factors $a^{r + 1} +^{n} C_{r + 1} (1)^{n - r - 1} a^{r + 1}$
Coefficient of $a^{r - 1} + a^{r + 1} = 2 a^{r}$
$=^{n} C_{r - 1} (1)^{n - r + 1} (a)^{r - 1} +^{n} C_{r + 1} (1)^{n - r - 1} a^{r + 1}$
$=^{n} C_{r} (a)^{r}$
$=^{n} C_{r - 1} a^{r - 1} +^{n} C_{r + 1} a^{r + 1} = 2^{n} C_{r} a^{r}$
$=^{n} C_{r - 1} (\frac{1}{a}) +^{n} C_{r + 1} a =^{2 n} C_{r}$

$= \frac{n!}{(n - r + 1)! \times (r - 1)!} (\frac{1}{a}) + \frac{n!}{(n - r - 1)! \times (r + 1)!} = \frac{2 n!}{(n - r)! r!}$

$= \frac{1}{a} (\frac{1}{n - r + 1}!) + \frac{a}{(n - r - 1)! r (r + 1)} = \frac{2}{(n - r)! \times r}$

$= \frac{1}{a} (\frac{1}{(n - r + 1) (n - r)}) + \frac{a}{r (r + 1)} = \frac{2}{(n - r) r}$

$= \frac{1}{(n - r + 1) (n - r)} + \frac{a^{2}}{r (r + 1)} = \frac{2 a}{(n - r) r}$

$r (r + 1) + a^{2} (n - r + 1) (n - r) = 2 a (n - r + 1) (r + 1)$
Simplify
$= n^{2} - n (4 r + 1) + 4 r^{2} - 2 = 0$

Suggest Corrections

Similar questions

If the coefficients of $a^{r - 1}, a^{r}$ and $a^{r + 1}$ in the binomial expansion $(1 + a)^{n}$ are in the arithmetic progression, prove that $n^{2} - n (4 r + 1) + 4 r^{2} - 2 = 0$

The 2nd, 3rd and 4th terms in the expansion of $(x + y)^{n}$ are 240, 720 and 1080, respectively. Find the values of x, y and n.

If the coefficients of $x^{r - 1}, x^{r}$ and $x^{r + 1}$ in the binomial expansion of $(1 + x)^{n}$ are in AP, prove that
$n^{2} - n (4 r + 1) + 4 r^{2} - 2 = 0$