If L=limn→∞ 1+3−11+3−21+3−41+3−8… 1+3−2n, then

1−3−1L=limn→∞ 1−3−11+3−11+3−21+3−4…1+3−2n 23L=limn→∞ 1−3−21+3−21+3−4…1+3−2n=limn→∞ 1−3−221+3−4…1+3−2n=limn→∞ 1−3−2n+1=1⇒L=3/2

If L=limn→∞ 1+3−11+3−21+3−41+3−8… 1+3−2n, then

1−3−1L=limn→∞ 1−3−11+3−11+3−21+3−4…1+3−2n 23L=limn→∞ 1−3−21+3−21+3−4…1+3−2n=limn→∞ 1−3−221+3−4…1+3−2n=limn→∞ 1−3−2n+1=1⇒L=3/2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $L = lim_{n \to \infty} (1 + 3^{- 1}) (1 + 3^{- 2}) (1 + 3^{- 4}) (1 + 3^{- 8}) \dots$ $(1 + 3^{- 2^{n}}),$ then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$L = \frac{2}{3}$

$L = 2$

$L = \frac{3}{2}$

$L = 1$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} (1 - 3^{- 1}) L = lim_{n \to \infty} (1 - 3^{- 1}) (1 + 3^{- 1}) (1 + 3^{- 2}) (1 + 3^{- 4}) \dots (1 + 3^{- 2^{n}}) \\ \frac{2}{3} L = lim_{n \to \infty} (1 - 3^{- 2}) (1 + 3^{- 2}) (1 + 3^{- 4}) \dots (1 + 3^{- 2^{n}}) \\ = lim_{n \to \infty} (1 - 3^{- 2^{2}}) (1 + 3^{- 4}) \dots (1 + 3^{- 2^{n}}) \\ = lim_{n \to \infty} (1 - 3^{- 2^{n + 1}}) = 1 \\ \Rightarrow L = 3 / 2 \end{array}$