Question

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

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

Answer 1

Answer: (A)

$x=\frac{3}{4\cdot 8}+\frac{3\cdot 5}{4\cdot 8\cdot 12}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12\cdot 16}+\dots \dots .$
Let, $(1+y{)}^n=1+ny+\frac{n(n-1)}{2!}{y}^2+\frac{n(n-1)(n-2)}{3!}+\frac{n(n-1)(n-2)(n-3)}{4!}{y}^4+\dots .$
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,
$x=(1+y{)}^n-1-ny={\left(1-\frac{1}{2}\right)}^{-1/2}-1-\left(-\frac{1}{2}\right)\left(\frac{-1}{2}\right)$
$\Rightarrow \sqrt{2}-1-\frac{1}{4}=\sqrt{2}-\frac{5}{4}$
Now, $2x^2=\left\{2+\frac{25}{16}-\frac{5}{\sqrt{2}}\right]=4+\frac{25}{8}-5\sqrt{2}$
and $5x=5\sqrt{2}-\frac{25}{4}$
So, $2x^2+5x=4+\frac{25}{8}-5\sqrt{2}+5\sqrt{2}-\frac{25}{4}$
$=\frac{32+25-50}{8}=\frac{7}{8}$