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If $\alpha=\frac{5}{2 ! \times 3}+\frac{5 \times 7}{3 ! \times 3^2}+\frac{5 \times 7 \times 9}{4 ! \times 3^3}+\ldots .$, then $\alpha^2+4 \alpha=$
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2530 Upvotes
Verified Answer
The correct answer is:
23
Since,
$$
\begin{aligned}
(1+x)^n=1+n x+ & \frac{n(n-1)}{2 !} x^2+\frac{n(n-1)(n-2)}{3 !} x^3 \\
& +\frac{n(n-1)(n-2)(n-3)}{4 !} x^4+\ldots
\end{aligned}
$$
On comparing
$\frac{n(n-1)}{2 !} x^2=\frac{5}{2 ! \times 3}$ and $\frac{n(n-1)(n-2)}{3 !} x^3=\frac{5 \times 7}{3 ! \times 3^2}$
So, $\quad x=-\frac{2}{3}$ and $n=-3 / 2$
$$
\begin{array}{ll}
\therefore(1-2 / 3)^{-3 / 2}=1+1+\alpha & \Rightarrow 3^{3 / 2}=2+\alpha \\
\Rightarrow \alpha^2+4 \alpha+4=27 & \Rightarrow \alpha^2+4 \alpha=23
\end{array}
$$
$$
\begin{aligned}
(1+x)^n=1+n x+ & \frac{n(n-1)}{2 !} x^2+\frac{n(n-1)(n-2)}{3 !} x^3 \\
& +\frac{n(n-1)(n-2)(n-3)}{4 !} x^4+\ldots
\end{aligned}
$$
On comparing
$\frac{n(n-1)}{2 !} x^2=\frac{5}{2 ! \times 3}$ and $\frac{n(n-1)(n-2)}{3 !} x^3=\frac{5 \times 7}{3 ! \times 3^2}$
So, $\quad x=-\frac{2}{3}$ and $n=-3 / 2$
$$
\begin{array}{ll}
\therefore(1-2 / 3)^{-3 / 2}=1+1+\alpha & \Rightarrow 3^{3 / 2}=2+\alpha \\
\Rightarrow \alpha^2+4 \alpha+4=27 & \Rightarrow \alpha^2+4 \alpha=23
\end{array}
$$
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