The coefficient of $x^{28}$ in the expansion of $(1 + x^{3} - x^{6})^{30}$ is

1

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

0

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

^{30} C_{6}

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

^{30} C_{3}

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

Open in App

Solution

The correct option is B $0$
$(1 + x^{3} - x^{6})^{30}$
The general term will be
$=^{30} C_{r} (x^{3} - x^{6})^{r}$
$=^{30} C_{r}^{r} C_{k} x^{3 r - 3 k} x^{6 k}$
$=^{30} C_{r}^{r} C_{k} x^{3 r - 3 k}$
For $x^{28}$
$3 (r - k) = 28$
$r - k = \frac{28}{3}$
Now, $r$ and $k$ both are whole numbers.
Hence in the above expansion, the term of $x^{28}$ does not exist.

Suggest Corrections

Similar questions

x^{28}

(1 + x^{3} - x^{6})^{30}

x^{9}

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

(1 + x + x^{2} + x^{3} + . . . . .

\infty)^{3}

x^{3}

x^{n}

\frac{1}{(1 - x) (3 - x)}

Coefficient of $x^{3}$ in the expansion of ${(\frac{1 + x}{1 - x})}^{2}$ , |x| < 1 is

Join BYJU'S Learning Program