Question

In the expansion of ${{\left( 1+\alpha x \right)}^{n}},n\in \mathbb{N}$, the coefficient of $x$ and ${{x}^{2}}$ are 8 and 24 respectively, then
[a] $\alpha =2,n=4$
[b] $\alpha =4,n=2$
[c] $\alpha =2,n=6$
[d] None of these.

Answer 1

Hint: Use the fact that the expansion of the expression ${{\left( x+y \right)}^{n}}$ is given by ${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}$. Hence find the coefficient of $x$and ${{x}^{2}}$ in the expansion of ${{\left( 1+\alpha x \right)}^{n}}$. Equate the coefficient to 8 and 24 and hence form two equations in $\alpha$ and n. Solve the system to find the value of $\alpha$ and n. Alternatively, assume that ${{\left( 1+\alpha x \right)}^{n}}=1+8x+24{{x}^{2}}+{{a}_{3}}{{x}^{3}}+\cdots$
Differentiate both sides, with respect x and put x = 0 and hence form an equation in $\alpha$ and $n$. Again differentiate with respect to x and put x =0 and hence form another equation in $\alpha$ and $n$. Solve the system and hence find the value of $\alpha$ and $n$.

Complete step-by-step anwer:

We know that ${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}$
Put x = 1 and $y=\alpha x$, we get
${{\left( 1+\alpha x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}\left( \alpha x \right){{+}^{n}}{{C}_{2}}{{\left( \alpha x \right)}^{2}}+\cdots$
Hence, we have
Coefficient of x in the expansion of ${{\left( 1+\alpha x \right)}^{n}}$ is equal to $^{n}{{C}_{1}}\alpha =n\alpha$
Coefficient of ${{x}^{2}}$ in the expansion of ${{\left( 1+\alpha x \right)}^{n}}$ is equal to $^{n}{{C}_{2}}{{\alpha }^{2}}=\dfrac{n\left( n-1 \right)}{2}{{\alpha }^{2}}$
Hence, we have
$\begin{aligned} & n\alpha =8\text{ }\left( i \right) \\\ & \dfrac{n\left( n-1 \right)}{2}{{\alpha }^{2}}=24\text{ }\left( ii \right) \\\ \end{aligned}$
Dividing equation (ii) by equation (i), we get
$\dfrac{n-1}{2}\alpha =3$
Also, from equation (i), we have
$\alpha =\dfrac{8}{n}$
Hence, we have
$\begin{aligned} & \dfrac{n-1}{2}\times \dfrac{8}{n}=3 \\\ & \Rightarrow \dfrac{n-1}{n}\left( 4 \right)=3 \\\ \end{aligned}$
Multiplying both sides by n, we get
4n-4 = 3n
Subtraction 3n from both sides, we get
n-4 = 0
Adding 4 on both sides, we get
n =4.
Substituting the value of n in equation (i), we get
$4\alpha =8$
Dividing both sides by 4, we get
$\alpha =2$
Hence option [a] is correct.

Note: Alternative Solution:
Let ${{\left( 1+\alpha x \right)}^{n}}=1+8x+24{{x}^{2}}+{{a}_{3}}{{x}^{3}}+\cdots$
Differentiating both sides with respect to x, we get
$n\alpha {{\left( 1+\alpha x \right)}^{n-1}}=8+48x+3{{a}_{3}}{{x}^{2}}+\cdots$
Put x =0, we get
$n\alpha =8$, which is the same as equation (i).
Again differentiating, we get
$n\left( n-1 \right){{\alpha }^{2}}{{\left( 1+\alpha x \right)}^{n-2}}=48+6{{a}_{3}}x+\cdots$
Put x =0, we get
$n\left( n-1 \right){{\alpha }^{2}}=48$, which is the same as equation (ii).
Hence following a similar procedure as above, we have $\alpha =2,n=4$
Hence option [d] is correct.
[2] Verification:
We know that ${{\left( 1+x \right)}^{4}}=1+4x+6{{x}^{2}}+4{{x}^{3}}+{{x}^{4}}$
Hence, we have
${{\left( 1+2x \right)}^{4}}=1+8x+24{{x}^{2}}+32{{x}^{3}}+16{{x}^{4}}$
Hence, the coefficient of x is 8 and the coefficient of ${{x}^{2}}$ is 24
Hence our answer is verified to be correct.