If [2 ^2 -5 =1] has exactly 7 solutions in the interval [[0, (n )/(2)]], for the least val…

Question

If $$2 \tan ^2 \theta-5 \sec \theta=1$$ has exactly 7 solutions in the interval $$\left[0, \frac{n \pi}{2}\right]$$, for the least value of $$n \in \mathbf{N}$$, then $$\sum_\limits{k=1}^n \frac{k}{2^k}$$ is equal to:

Accepted Answer

$$\begin{aligned}
& 2 \tan ^2 \theta-5 \sec \theta-1=0 \
& \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \
& \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \
& \Rightarrow \sec \theta=-\frac{1}{2}, 3 \
& \Rightarrow \cos \theta=-2, \frac{1}{3} \
& \Rightarrow \cos \theta=\frac{1}{3}
\end{aligned}$$

For 7 solutions $$\mathrm{n}=13$$

$$\begin{aligned}
& \text { So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text { (say) } \
& \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots .+\frac{13}{2^{13}} \
& \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}} \
& \Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}
\end{aligned}$$

Solutions · 1