If the coefficient of x in (x^ 2 + x )^ 5 is 270, then = - Vidyadip

MCQ

If the coefficient of $x$ in $\Big(\text{x}^{2}+\frac{\lambda}{\text{x}}\Big)^{5}$ is $270,$ then $\lambda=$

✓
$3$
B
$4$
C
$5$
D
None of these.

✓

Answer

Correct option: A.

$3$

The coefficient of $x$ in the given expansion where $x$ occurs at the $(r + 1)^{th}$ term.
We have,
${^\text{15}}\text{C}_{\text{r}}(\text{x}^{2})^{5-\text{r}}\ \Big(\frac{\lambda}{\text{x}}\Big)^{\text{r}}$
$={^\text{15}}\text{C}_{\text{r}}\ \lambda^{\text{r}}\ \text{x}^{10-2\text{r}-\text{r}}$
For it to contain $x$, we must have
$10-3\text{r}=1$
$\Rightarrow \text{r}=3$
Coefficient of $x$ in the given expansion,
$={^\text{15}}\text{C}_{\text{3}}\ \lambda^{\text{3}}$
$=10\lambda^{3}$
Now, we have
$10\lambda^{3}=270$
$\Rightarrow \lambda^{3}=27$
$\Rightarrow \lambda=3$

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