If α= log e √ cot (π/12), then find the value of (( displaystyle∑k=0∞ e-2 k α/ displaystyle∑k=0∞(-1)k e-2 k x))2

Question

If α= log e √ cot (π/12), then find the value of (( displaystyle∑k=0∞ e-2 k α/ displaystyle∑k=0∞(-1)k e-2 k x))2 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Expression =(1+ e -2 α+ e -4 α+ ldots ldots ∞/1- e -2 α+ e -4 α+ ldots ldots ∞)=(1+ e -2 α/1- e -2 α)=(( e 2 α+1/ e 2 α-1)) As, e α=√ cot (π/12) ⇒ e 2 α=(2+√3). ∴ Expression =√3.

Q. If $α = lo g_{e} cot \frac{π}{12}$ , then find the value of $⎝ ⎛ \frac{k = 0 \sum \infty e ^{- 2 k α}}{k = 0 \sum \infty ( - 1 ) ^{k} e ^{- 2 k x}} ⎠ ⎞^{2}$

Expression $= \frac{1 + e ^{- 2 α} + e ^{- 4 α} + \dots\dots \infty}{1 - e ^{- 2 α} + e ^{- 4 α} + \dots\dots \infty} = \frac{1 + e ^{- 2 α}}{1 - e ^{- 2 α}} = (\frac{e ^{2 α} + 1}{e ^{2 α} - 1})$
As, $e^{α} = cot \frac{π}{12} \Rightarrow e^{2 α} = (2 + 3)$ .
$∴$ Expression $= 3$ .

Q. If α=loge​cot12π​​, then find the value of ⎝⎛​k=0∑∞​(−1)ke−2kxk=0∑∞​e−2kα​⎠⎞​2

Expression =1−e−2α+e−4α+……∞1+e−2α+e−4α+……∞​=1−e−2α1+e−2α​=(e2α−1e2α+1​) As, eα=cot12π​​⇒e2α=(2+3​). ∴ Expression =3​.

Q. If $α = lo g_{e} cot \frac{π}{12}$ , then find the value of $⎝ ⎛ \frac{k = 0 \sum \infty e ^{- 2 k α}}{k = 0 \sum \infty ( - 1 ) ^{k} e ^{- 2 k x}} ⎠ ⎞^{2}$

Expression $= \frac{1 + e ^{- 2 α} + e ^{- 4 α} + \dots\dots \infty}{1 - e ^{- 2 α} + e ^{- 4 α} + \dots\dots \infty} = \frac{1 + e ^{- 2 α}}{1 - e ^{- 2 α}} = (\frac{e ^{2 α} + 1}{e ^{2 α} - 1})$
As, $e^{α} = cot \frac{π}{12} \Rightarrow e^{2 α} = (2 + 3)$ .
$∴$ Expression $= 3$ .