Question

If $x = \frac{3}{4.8} + \frac{3.5}{4.8.12}+\frac{3.5.7}{4.8.12.16} + ..., $ then $2x^{2} + 5x = $

• A. $\frac{7}{8}$
• B. 7
• C. $\frac{7}{16}$
• D. $\frac{7}{4}$

Answer 1

Answer: (A)

$x=\frac{3}{4 \cdot 8}+\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots .$
Let, $(1+y)^{n}=1+n y+ \frac{n(n-1)}{2 !} y^{2}+\frac{n(n-1)(n-2)}{3 !}+\frac{n(n-1)(n-2)(n-3)}{4 !} y^{4}+\ldots .$
If we compare with third term on words, the
$\frac{n(n-1)}{2 !} y^{2}=\frac{3}{1.8}, \frac{n(n-1)(n-2)}{n !}$
$y^{3}=\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}$
and $\frac{n(n-1)(n-2)(n-3)}{4 !} y^{4}$
$=\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}$
On solving, we are getting. So,
$x=(1+y)^{n} -1-n y=\left(1-\frac{1}{2}\right)^{-1 / 2}-1-\left(-\frac{1}{2}\right)\left(\frac{-1}{2}\right)$
$\Rightarrow \sqrt{2}-1-\frac{1}{4} =\sqrt{2}-\frac{5}{4}$
Now, $2 x^{2} =\left\{2+\frac{25}{16}-\frac{5}{\sqrt{2}}\right]=4+\frac{25}{8}-5 \sqrt{2}$
and $5 x =5 \sqrt{2}-\frac{25}{4}$
So, $2 x^{2}+5 x =4+\frac{25}{8}-5 \sqrt{2}+5 \sqrt{2}-\frac{25}{4}$
$=\frac{32+25-50}{8}=\frac{7}{8}$