If 2 tan^2θ -5 secθ =1 has exactly 7 solutions in the interval [0, nπ 2], for the least value of n∈ N then Σ _k=1^n k 2^k is equal to :

Given,

$$2{\mathrm{tan}}^{2}\theta -5sec\theta -1=0$$2tan2θ−5secθ−1=0

$$\Rightarrow 2({\mathrm{sec}}^{2}\theta -1)-5sec\theta -1=0$$⇒2(sec2θ−1)−5secθ−1=0

$$\Rightarrow 2{\mathrm{sec}}^{2}\theta -5sec\theta -3=0$$⇒2sec2θ−5secθ−3=0

$$\Rightarrow 2{\mathrm{sec}}^{2}\theta -6sec\theta +sec\theta -3=0$$⇒2sec2θ−6secθ+secθ−3=0

$$\Rightarrow 2sec\theta (\mathrm{sec}\theta -3)+1(sec\theta -3)=0$$⇒2secθ(secθ−3)+1(secθ−3)=0

$$\Rightarrow (2sec\theta +1)(sec\theta -3)=0$$⇒(2secθ+1)(secθ−3)=0

$$\Rightarrow sec\theta =-\frac{1}{2},3$$⇒secθ=−21​,3

$$\Rightarrow cos\theta =-2,\frac{1}{3}$$⇒cosθ=−2,31​

We know that, $-1\leq cos\theta \leq 1$−1≤cosθ≤1

$$\Rightarrow cos\theta =\frac{1}{3}$$⇒cosθ=31​

For $7$7 solutions $n=13$n=13

So, $\displaystyle \sum _{k=1}^{13}\frac{k}{{2}^{k}}=S$k=1∑13​2kk​=S (say)

$$\Rightarrow S=\frac{1}{2}+\frac{2}{{2}^{2}}+\frac{3}{{2}^{3}}+\ldots .+\frac{13}{{2}^{13}}$$⇒S=21​+222​+233​+….+21313​

$$\Rightarrow \frac{1}{2}S=\frac{1}{{2}^{2}}+\frac{1}{{2}^{3}}+\ldots ..+\frac{12}{{2}^{13}}+\frac{13}{{2}^{14}}$$⇒21​S=221​+231​+…..+21312​+21413​

$$\Rightarrow S-\frac{S}{2}=(\frac{1}{2}+\frac{2}{{2}^{2}}+\frac{3}{{2}^{3}}+\ldots .+\frac{13}{{2}^{13}})-(\frac{1}{{2}^{2}}+\frac{1}{{2}^{3}}+\ldots ..+\frac{12}{{2}^{13}}+\frac{13}{{2}^{14}})$$⇒S−2S​=(21​+222​+233​+….+21313​)−(221​+231​+…..+21312​+21413​)

$$\Rightarrow \frac{S}{2}=(\frac{1}{2}+\frac{1}{{2}^{2}}+\frac{3}{{2}^{3}}+...+\frac{13}{{2}^{13}})-\frac{13}{{2}^{14}}$$⇒2S​=(21​+221​+233​+...+21313​)−21413​

$$\Rightarrow \frac{S}{2}=\frac{1}{2}\cdot \frac{1-\frac{1}{{2}^{13}}}{1-\frac{1}{2}}-\frac{13}{{2}^{14}}$$⇒2S​=21​⋅1−21​1−2131​​−21413​

$$\Rightarrow \frac{S}{2}=\frac{1}{2}\cdot \frac{\frac{{2}^{13}-1}{{2}^{13}}}{\frac{1}{2}}-\frac{13}{{2}^{14}}$$⇒2S​=21​⋅21​213213−1​​−21413​

$$\Rightarrow S=2\cdot (\frac{{2}^{13}-1}{{2}^{13}})-\frac{13}{{2}^{13}}$$⇒S=2⋅(213213−1​)−21313​

$$\Rightarrow S=(\frac{{2}^{14}-2-13}{{2}^{13}})$$⇒S=(213214−2−13​)

$$\Rightarrow S=(\frac{{2}^{14}-15}{{2}^{13}})$$⇒S=(213214−15​)