JEE Main 2021 — Mathematics Trigonometry

Given $\displaystyle {S}_{k}=\sum _{r=1}^{k}{\mathrm{tan}}^{-1}(\frac{{6}^{r}}{{2}^{2r+1}+{3}^{2r+1}})$Sk​=r=1∑k​tan−1(22r+1+32r+16r​)

$$\Rightarrow {S}_{k}=\sum _{r=1}^{k}{\mathrm{tan}}^{-1}(\frac{{2}^{r}\cdot {3}^{r}}{{2}^{2r}\cdot 2+{3}^{2r}\cdot 3})$$⇒Sk​=r=1∑k​tan−1(22r⋅2+32r⋅32r⋅3r​)

Divide by ${3}^{2r}$32r we get,

$${S}_{k}=\sum _{r=1}^{k}{\mathrm{tan}}^{-1}(\frac{{(\frac{2}{3})}^{r}}{{(\frac{2}{3})}^{2r}.2+3})$$Sk​=r=1∑k​tan−1((32​)2r.2+3(32​)r​)

$${S}_{k}=\sum _{r=1}^{k}{\mathrm{tan}}^{-1}(\frac{{(\frac{2}{3})}^{r}}{3({(\frac{2}{3})}^{2r+1}+1)})$$Sk​=r=1∑k​tan−1(3((32​)2r+1+1)(32​)r​)

Let ${(\frac{2}{3})}^{r}=t$(32​)r=t

$${S}_{k}=\sum _{r=1}^{k}{\mathrm{tan}}^{-1}(\frac{\frac{t}{3}}{1+\frac{2}{3}{t}^{2}})$$Sk​=r=1∑k​tan−1(1+32​t23t​​)

$${S}_{k}=\sum _{r=1}^{k}{\mathrm{tan}}^{-1}(\frac{t-\frac{2t}{3}}{1+t.\frac{2t}{3}})$$Sk​=r=1∑k​tan−1(1+t.32t​t−32t​​)

$${S}_{k}=\sum _{r=1}^{k}({\mathrm{tan}}^{-1}(t)-{\mathrm{tan}}^{-1}(\frac{2t}{3}))$$Sk​=r=1∑k​(tan−1(t)−tan−1(32t​))

$${S}_{k}=\sum _{r=1}^{k}({\mathrm{tan}}^{-1}{(\frac{2}{3})}^{r}-{\mathrm{tan}}^{-1}{(\frac{2}{3})}^{r+1})$$Sk​=r=1∑k​(tan−1(32​)r−tan−1(32​)r+1)

$${S}_{k}=({\mathrm{tan}}^{-1}(\frac{2}{3})-{\mathrm{tan}}^{-1}{(\frac{2}{3})}^{2})+({\mathrm{tan}}^{-1}{(\frac{2}{3})}^{2}-{\mathrm{tan}}^{-1}{(\frac{2}{3})}^{3})+...+({\mathrm{tan}}^{-1}{(\frac{2}{3})}^{k}-{\mathrm{tan}}^{-1}{(\frac{2}{3})}^{k+1})$$Sk​=(tan−1(32​)−tan−1(32​)2)+(tan−1(32​)2−tan−1(32​)3)+...+(tan−1(32​)k−tan−1(32​)k+1)

$${S}_{k}={\mathrm{tan}}^{-1}(\frac{2}{3})-{\mathrm{tan}}^{-1}{(\frac{2}{3})}^{k+1}$$Sk​=tan−1(32​)−tan−1(32​)k+1

Then, ${S}_{\infty }=\underset{k\rightarrow \infty }{\mathrm{lim}}({\mathrm{tan}}^{-1}(\frac{2}{3})-{\mathrm{tan}}^{-1}{(\frac{2}{3})}^{k+1})$S∞​=k→∞lim​(tan−1(32​)−tan−1(32​)k+1)

$$={\mathrm{tan}}^{-1}(\frac{2}{3})-{\mathrm{tan}}^{-1}(0)$$=tan−1(32​)−tan−1(0)

$$\therefore {S}_{\infty }={\mathrm{tan}}^{-1}(\frac{2}{3})={\mathrm{cot}}^{-1}(\frac{3}{2}),(\because {\mathrm{tan}}^{-1}x={\mathrm{cot}}^{-1}(\frac{1}{x}))$$∴S∞​=tan−1(32​)=cot−1(23​),(∵tan−1x=cot−1(x1​))