The coefficient of $x^{9}$ in $(1 + 9 x + 27 x^{2} + 27 x^{3})^{6}$ is

^{18} C_{9} {.3}^{9}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

^{18} C_{8} {.3}^{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{18} C_{10} {.3}^{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{8} C_{9} {.3}^{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $^{18} C_{9} {.3}^{9}$
$(1 + 9 x + 27 x^{2} + 27 x^{3})^{6}$
$= ((1 + 3 x)^{3})^{6}$
$= (1 + 3 x)^{18}$
$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$
Applying to the above question we get
$T_{r + 1} =^{18} C_{r} 1^{n - r} (3 x)^{r}$
$=^{18} C_{r} (3 x)^{r}$ ...(i)
For the coefficient of $x^{9}$
$r = 9$
Substituting in (i) we get
$T_{10} =^{18} C_{9} 3^{9} x^{9}$

Suggest Corrections

Similar questions

x^{9}

\to (1 + x) ((1 + x^{2}) (1 + x^{3}) . . . . . . (1 + x^{100})) ?

(1 + a x)^{n} = 1 + 9 x + 27 x^{2} + \dots \dots

(a, n) =

x^{9}

(x + 2) (x + 4) (x + 8) . . . . . . . (x + 1024)

x^{7}

(1 - x^{4}) (1 + x)^{9}

x^{9}

{(x^{2} - \frac{1}{3 x})}^{9}

Join BYJU'S Learning Program