Question

Evaluate ${\sum }_{r=1}^n\left[{\sum }_{k=1}^{r}k\right]{\left[{\log }_{1/2}\sqrt{(4x-x^2)}\right]}^r$. Find $x$ for which summation is a finite number as $n\to \infty$

• A. $x\in (0,2+\frac{\sqrt{15}}{2})$
• B. $x\in (0,2-\frac{\sqrt{15}}{2})$
• C. $x\in (0,-2+\frac{\sqrt{15}}{2})$
• D. $x\in (0,-2-\frac{\sqrt{15}}{2})$

Answer 1

The correct option is A $x\in (0,2+\frac{\sqrt{15}}{2})$

We have,
${\sum }_{r=1}^n\left[{\sum }_{k=1}^{r}k\right]{\left[{\log }_{1/2}\sqrt{(4x-x^2)}\right]}^r$
$={\sum }_{r=1}^n\frac{r(r+1)}{2}{\left[{\log }_{1/2}\sqrt{(4x-x^2)}\right]}^r$
$={\sum }_{r=1}^n\frac{r(r+1)}{2}{A}^r$ [where $A={\log }_{1/2}\sqrt{(4x-x^2)}$]
$=\frac{1}{2}\sum _{r=1}^{n}r(r+1)A^r$
$=\frac{1}{2}\left[\right(1.2)A+(2.3)A^2+(3.4)A^3+....+n(n+1\left)A^n\right]$
$S=\frac{1}{2}{S}_1$ (say) $\cdots \left(1\right)$
$\therefore S_1=\left(1.2\right)A+\left(2.3\right)A^2+\left(3.4\right)A^3+....+(n-1)n{A}^{n-1}+n(n+1)A^n$
$\Rightarrow S_1=2A+6A^2+12A^3+.....+(n-1)n{A}^{n-1}+n(n+1)A^n$
$\therefore A{S}_1=0+2A^2+6A^3+....+(n-1)n{A}^n+n(n+1)A^{n+1}$
Subtracting, we get,
$(1-A)S_1=2A+4A^2+6A^3+....+2n{A}^n-n(n+1)A^{n+1}$
$\Rightarrow (1-A)S_1=2[A+2A^2+3A^3+...+n{A}^n]-n(n+1)A^{n+1}$
$=2S_2-n(n+1)A^{n+1}$ (say) $\cdots \left(2\right)$
$\therefore S_2=A+2A^2+3A^3+....+(n-1)A^{n-1}+n{A}^n$
from $\left(2\right)$,
$(1-A)S_1=\frac{2A(1-A^n)}{(1-A{)}^2}-\frac{2n{A}^{n+1}}{(1-A)}-n(n+1)A^{n+1}$
$\therefore S_1=\frac{2A(1-A^n)}{(1-A{)}^3}-\frac{2n{A}^{n+1}}{(1-A{)}^2}-\frac{n(n+1)A^{n+1}}{(1-A)}$
now from $\left(1\right)$,
$S=\frac{1}{2}{S}_1$
$\therefore S=\frac{A(1-A^n)}{(1-A{)}^3}-\frac{n{A}^{n+1}}{(1-A{)}^2}-\frac{n(n+1)A^{n+1}}{2(1-A)}$
Now, for $S$ to be finite we must have $|A|<1$
$S_{\infty }=\frac{A(1-0)}{(1-A{)}^3}-0-0$
$=\frac{A}{(1-A{)}^3}=\frac{{\log }_{1/2}\sqrt{(4x-x^2)}}{(1-{\log }_{1/2}\sqrt{(4x-x^2)}{)}^3}=finite$
$(\because A<1)$
i.e ${\log }_{1/2}\sqrt{(4x-x^2)}<1$
$\Rightarrow \sqrt{4x-x^2}>\frac{1}{2}$
$\Rightarrow 4x-x^2>\frac{1}{4}$
$\Rightarrow 4x^2-16x+1<0$
$\therefore 2-\frac{\sqrt{15}}{2}<x<2+\frac{\sqrt{15}}{2}\cdots \left(3\right)$
and $4x-x^2>0$
$\Rightarrow x^2-4x<0$
$\Rightarrow x(x-4)<0$
$\therefore 0<x<4\cdots \left(4\right)$
combining $\left(3\right)$ and $\left(4\right)$ we get
$x\in (0,2+\frac{\sqrt{15}}{2})$