उत्तर (4)

समाधान

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

7 हल के लिए $n=13$

इसलिए, $\sum _{k=1}^{13} \frac{k}{2^{k}}=S$ (मान लीजिए)

$S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots+\frac{13}{2^{13}}$

$\frac{1}{2} S=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}}$

$\Rightarrow \frac{S}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow S=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}$