In the following expansion find which is the greatest term:
${(x - y)}^{30}$ when $x = 11, y = 4$ .

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Solution

General term in the binomial expansion of ${(a + b)}^{n}$ is

$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$ and

$T_{r + 1} > T_{r}$

i.e. $^{n} C_{r} \cdot a^{n - r} \cdot b^{r} >^{n} C_{r - 1} \cdot a^{n - r + 1} \cdot b^{r - 1}$

$\Rightarrow \frac{n - r + 1}{r} \times b > a$

$\Rightarrow \frac{n - r + 1}{r} > \frac{a}{b}$

$\Rightarrow n - r + 1 > \frac{r a}{b}$

$\Rightarrow n + 1 > r (1 + \frac{a}{b})$

$\Rightarrow r < b (\frac{n + 1}{a + b})$

$\Rightarrow r < \frac{31 \times 4}{15}$

$\Rightarrow r < 8.266$
$\Rightarrow r = 8$

$∴ T_{9}$ is greatest

$T_{9} =^{30} C_{8} {(11)}^{22} {(- 4)}^{8}$

$=^{30} C_{8} \cdot 11^{22} \times 4^{8}$

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