Question

If $\alpha=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^{2}}+\frac{5 \cdot 7 \cdot 9}{4 ! 3^{3}}+\ldots, $ then $\alpha^{2}+4 \alpha$ is equal to

• A. 21
• B. 23
• C. 25
• D. 27

Answer 1

Answer: (B)

Given that, $\alpha=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^{2}}+\frac{5 \cdot 7 \cdot 9}{4 ! 3^{3}}+\ldots\,\,\,...(i)$
We know that,
$(1+x)^{n}= 1+\frac{n x}{1 !}+\frac{n(n-1)}{2 !} x^{2} $
$+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots\,\,\, . . .(ii)$
On comparing Eqs. (i) and (ii), with respect to factorial
$n(n-1) x^{2} =\frac{5}{3} \,\,\, \ldots (iii)$
$n(n-1)(n-2) x^{3} =\frac{5 \cdot 7}{3^{2}} \,\,\, \ldots(iv)$
and
$n(n-1)(n-2)(n-3) x^{4}=\frac{5 \cdot 7 \cdot 9}{3^{3}} \,\,\,\ldots( v )$
On dividing Eq. (iv) by (iii) and Eq. (v) by (iv), we get
$(n-2) x=\frac{7}{3}\,\,\,...(vi)$
and $(n-3) x=3\,\,\,\cdots(vii)$
Again, dividing Eq. (vi) by (vii), we get
$\frac{n-2}{n-3} =\frac{7}{9} $
$\Rightarrow 9 n-18 =7 n-21 $
$\Rightarrow 2 n =-3 $
$\Rightarrow n =-\frac{3}{2} $
On putting the value of $n$ in Eq. (vi),
we get
$\left(-\frac{3}{2}-2\right) x=\frac{7}{3} $
$\Rightarrow x=-\frac{2}{3}$
$\therefore $ From Eq. (ii),
$\left(1-\frac{2}{3}\right)^{-3 / 2} =1+1+\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^{2}}+\ldots $
$ \Rightarrow 3^{3 / 2}-2 =\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^{2}}+\ldots$
$ \Rightarrow \alpha=3^{3 / 2}-2 \,\,\,$[from Eq. (i) ]
Now, $ \alpha^{2}+4 \alpha =\left(3^{3 / 2}-2\right)^{2}+4\left(3^{3 / 2}-2\right) $
$=27+4-4 \cdot 3^{3 / 2}+4 \cdot 3^{3 / 2}-8 $
$=23 $