Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Option D)

$log_{e}( \frac{4}{2} )$

Refer the attachment

Answer 2

Answer:

Sum of series 1/1*2 - 1/2*3 + 1/3*4 --------- upto is logₑ(4/e)

Step-by-step explanation:

solution :- We have 1/1*2 - 1/2*3 + 1/3*4 -------upto ∞

Now , we can write this like as

• = (1 - 1/2 ) - (1/2 -1/3) + (1/3 - 1/4) -----∞

• = 1 - (2 * 1/2 )+ (2* 1/3) - (2* 1/4 ) + -----∞

• = 2 ( 1 - 1/2 + 1/3 - 1/4 + .-------∞) - 1

Used logarithmic series

log(1+x) = x - x²/2 + x³/3 - x⁴/4

Similarly here x = 1

• ∴ 2 log (1 + 1) - 1

• = log (2)² - loge

• = logₑ(4/e)Answer