Question

If $x=\frac{2.5}{(2!)3}+\frac{\mathrm{2.5.7}}{(3!)3^2}+\frac{\mathrm{2.5.7.9}}{(4!)3^3}+....$ then $x^2+8x+8=$

• A. 108
• B. 100
• C. 27
• D. 23

Answer 1

The correct option is B 100

$x=\frac{2.5}{2!3}+\frac{\mathrm{2.5.7}}{3!3^2}+\frac{\mathrm{2.5.7.9}}{4!3^3}+.......$

Divide both sides by $2$ and multiply by $3$

$3\frac{x}{2}=\frac{3.5}{2!3}+\frac{\mathrm{3.5.7}}{3!3^2}+\frac{\mathrm{3.5.7.9}}{4!3^3}+.......$

Divide by $3$

$\frac{x}{2}=\frac{3.5}{2!}{\left(\frac{1}{3}\right)}^2+\frac{\mathrm{3.5.7}}{3!}{\left(\frac{1}{3}\right)}^3+\frac{\mathrm{3.5.7.9}}{4!3^4}+.....$

Adding $1+\frac{3}{5}$ on both sides.

$\frac{x}{2}+1+\frac{1}{11}=1+\frac{3}{11}\left(\frac{1}{3}\right)+\frac{3.5}{2!}{\left(\frac{1}{3}\right)}^2+\frac{\mathrm{3.5.7}}{3!}{\left(\frac{1}{3}\right)}^3+\frac{\mathrm{3.5.7.9}}{4!}{\left(\frac{1}{3}\right)}^4+.......$

here $p=3,y=2$ and $\frac{x}{y}=\frac{1}{3}$

$So,x=\frac{2}{3}\frac{x}{2}+2={(1-\frac{2}{3})}^{\frac{-3}{2}}={\left(\frac{1}{3}\right)}^{\frac{-3}{2}}\frac{x}{2}=3^{\frac{3}{2}}-2\frac{x}{2}=3\sqrt{3}-2So,x^2+8x+8=4{(3\sqrt{3}-2)}^2+2.8(3\sqrt{3}-2)+8=4(27+4-12\sqrt{3})+16(3\sqrt{3}-2)+8=4(31-12\sqrt{3})+48\sqrt{3}-32+8=124-48\sqrt{3}+48\sqrt{3}-24=100$