# Consider the expansion (3x^2 + (1/x))^6 How many terms does the expansion include? Find the constant term Show that the expansion has no terms involving x^5

Jan 29, 2018

Expansion of ${\left(3 {x}^{2} + \frac{1}{x}\right)}^{6}$ has seven terms. Constant term is$135$. For details see below.

#### Explanation:

An expansionof ${\left(a + b\right)}^{6}$ has seven terms and ${\left(3 {x}^{2} + \frac{1}{x}\right)}^{6}$ too has seven terms. This may be seen from the following.

The expansion of ${\left(a + b\right)}^{n}$ is ${C}_{0}^{n} {a}^{n} + {C}_{1}^{n} {a}^{n - 1} {b}^{1} + {C}_{2}^{n} {a}^{n - 2} {b}^{2} + \ldots . . + {C}_{r}^{n} {a}^{n - r} {b}^{r} + \ldots + {C}_{n}^{n} {b}^{n}$

Observe that ${\left(r + 1\right)}^{t h}$ term is ${C}_{r}^{n} {a}^{n - r} {b}^{r}$

hence ${\left(r + 1\right)}^{t h}$ term of ${\left(3 {x}^{2} + \frac{1}{x}\right)}^{6}$ is

${C}_{r}^{6} {\left(3 {x}^{2}\right)}^{6 - r} {\left(\frac{1}{x}\right)}^{r} = {C}_{r}^{6} {3}^{6 - r} {x}^{12 - 3 r}$

Observe that $r$ can take values only from $0$ to $6$, for which power of $x$ can be $12 , 9 , 6 , 3 , 0 , - 3 , - 6$,

Hence expansion does not have a term ${x}^{5}$.

Constant term means ${x}^{0}$, which is for $r = 4$ and it is

${C}_{4}^{6} {3}^{2} = \frac{6 \cdot 5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4} \cdot 9 = 135$