Question

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

• A. 21
• B. 23
• C. 25
• D. 27

Answer 1

Answer: (B)

Given that, $\alpha =\frac{5}{2!3}+\frac{5\cdot 7}{3!3^2}+\frac{5\cdot 7\cdot 9}{4!3^3}+\dots ...(i)$
We know that,
$(1+x{)}^n=1+\frac{nx}{1!}+\frac{n(n-1)}{2!}{x}^2$
$+\frac{n(n-1)(n-2)}{3!}{x}^3+\dots ...(ii)$
On comparing Eqs. (i) and (ii), with respect to factorial
$n(n-1)x^2=\frac{5}{3}\dots (iii)$
$n(n-1)(n-2)x^3=\frac{5\cdot 7}{3^2}\dots (iv)$
and
$n(n-1)(n-2)(n-3)x^4=\frac{5\cdot 7\cdot 9}{3^3}\dots (v)$
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
$\frac{n-2}{n-3}=\frac{7}{9}$
$\Rightarrow 9n-18=7n-21$
$\Rightarrow 2n=-3$
$\Rightarrow n=-\frac{3}{2}$
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),
${\left(1-\frac{2}{3}\right)}^{-3/2}=1+1+\frac{5}{2!3}+\frac{5\cdot 7}{3!3^2}+\dots$
$\Rightarrow 3^{3/2}-2=\frac{5}{2!3}+\frac{5\cdot 7}{3!3^2}+\dots$
$\Rightarrow \alpha =3^{3/2}-2$[from Eq. (i) ]
Now, ${\alpha }^2+4\alpha ={\left(3^{3/2}-2\right)}^2+4\left(3^{3/2}-2\right)$
$=27+4-4\cdot 3^{3/2}+4\cdot 3^{3/2}-8$
$=23$