Question

If $y=\frac{3}{4}+\frac{3.5}{4.8}+\frac{\mathrm{3.5.7}}{\mathrm{4.8.12}}+\dots \dots \dots \infty$ then

• A. $y^2-2y+5=0$
• B. $y^2+2y-7=0$
• C. $y^2-3y+4=0$
• D. $y^2+4y-6=0$

Answer 1

Answer: (B)

$y=\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12}+\dots \infty$
$y+1=1+\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12}+\dots \infty$
$=1+\left(\frac{-3}{2}\right)\left(\frac{-1}{2}\right)+\frac{\left(\frac{-3}{2}\right)\left(\frac{-3}{2}-1\right){\left(\frac{-1}{2}\right)}^2}{2!}+\frac{\left(\frac{-3}{2}\right)\left(\frac{-3}{2}-1\right)\left(\frac{-3}{2}-2\right){\left(\frac{-1}{2}\right)}^3}{3!}+\dots$
$={\left(1-\frac{1}{2}\right)}^{-3/2}={\left(\frac{1}{2}\right)}^{-3/2}=2^{3/2}=2\sqrt{2}$
$\therefore y+1=2\sqrt{2}$
$\Rightarrow (y+1{)}^2=8$
$\Rightarrow y^2+2y+1=8$
$\Rightarrow y^2+2y-7=0$