Question

If $\alpha =\frac{5}{2!3}+\frac{5.7}{3!3^2}+\frac{\mathrm{5.7.9}}{4!3^3}$,.... then find the value of ${\alpha }^2+4\alpha$.

Answer 1

Let, $\alpha =\frac{5}{2!3}+\frac{5.7}{3!3^2}+\frac{\mathrm{5.7.9}}{4!3^3}+........$

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

or, $\alpha =\frac{3.(3+2)}{2!3^2}+\frac{3.(3+2).(3+4)}{3!3^3}+\frac{3.(3+2).(3+4).(3+6)}{4!3^4}+........$

or, $\alpha =\{1+\frac{\frac{3}{2}}{1!}\left(\frac{2}{3}\right)+\frac{\frac{3}{2}.(\frac{3}{2}+1)2^2}{2!3^2}+\frac{\frac{3}{2}.(\frac{3}{2}+1).(\frac{3}{2}+2)2^3}{3!3^3}+\frac{\frac{3}{2}.(\frac{3}{2}+1).(\frac{3}{2}+2).(\frac{3}{2}+3)2^4}{4!3^4}+........\}-2$

or, $\alpha =\{1+\frac{\frac{3}{2}}{1!}\left(\frac{2}{3}\right)+\frac{\frac{3}{2}.(\frac{3}{2}+1)}{2!}{\left(\frac{2}{3}\right)}^2+\frac{\frac{3}{2}.(\frac{3}{2}+1).(\frac{3}{2}+2)}{3!}{\left(\frac{2}{3}\right)}^3+\frac{\frac{3}{2}.(\frac{3}{2}+1).(\frac{3}{2}+2).(\frac{3}{2}+3)}{4!}{\left(\frac{2}{3}\right)}^4+........\}-2$

or, $\alpha ={(1-\frac{2}{3})}^{-\frac{3}{2}}-2={\left(\frac{1}{3}\right)}^{-\frac{3}{2}}-2=3^{\frac{3}{2}}-2$.

Now, ${\alpha }^2+4\alpha =(3^3-4\times 3\sqrt{3}+4)+4(3\sqrt{3}-2)=27-4=23$