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If in the expansion of $(1+p x)^{n}, n \in N$, the coefficient of $x$ and $x^{2}$ are 8 and 24 , then
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Verified Answer
The correct answer is:
$n=4, p=2$
We have, $(1+p x)^{n}$
$$
\therefore \quad T_{r+1}={ }^{n} C_{r}(1)^{n-r}(p x)^{r} \Rightarrow T_{r+1}={ }^{n} C_{r} p^{r} x^{r}
$$
Coefficient of $x^{r}={ }^{n} C_{r} p^{r}$
Now, coefficient of $x={ }^{n} C_{1} p=8 \quad$ (put $r=1$ )
$\Rightarrow \quad n p=8 \quad \ldots$ (i)
Also, coefficient of $x^{2}={ }^{n} C_{2} p^{2}=24 \quad$ (put $\left.r=2\right)$
$\frac{n(n-1)}{2} p^{2}=24 \quad \text{[from Eq.(i)]}$
$\Rightarrow \quad \frac{n(n-1)}{2}\left(\frac{8}{n}\right)^{2}=24$
$\Rightarrow \quad \frac{64(n-1)}{2 n}=24 \Rightarrow 4(n-1)=3 n \Rightarrow n=4$
From Eq. (i), we get $p=2$
$$
\therefore \quad T_{r+1}={ }^{n} C_{r}(1)^{n-r}(p x)^{r} \Rightarrow T_{r+1}={ }^{n} C_{r} p^{r} x^{r}
$$
Coefficient of $x^{r}={ }^{n} C_{r} p^{r}$
Now, coefficient of $x={ }^{n} C_{1} p=8 \quad$ (put $r=1$ )
$\Rightarrow \quad n p=8 \quad \ldots$ (i)
Also, coefficient of $x^{2}={ }^{n} C_{2} p^{2}=24 \quad$ (put $\left.r=2\right)$
$\frac{n(n-1)}{2} p^{2}=24 \quad \text{[from Eq.(i)]}$
$\Rightarrow \quad \frac{n(n-1)}{2}\left(\frac{8}{n}\right)^{2}=24$
$\Rightarrow \quad \frac{64(n-1)}{2 n}=24 \Rightarrow 4(n-1)=3 n \Rightarrow n=4$
From Eq. (i), we get $p=2$
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