sir, in last question how is x less that 0.5 and greater that 0.44….i already read the concept in earlier blog but there also i was having the same confusion…
If you are not comfortable with calculation (tables, basic percentage-ratio conversion, squares, and so on), then you will NOT be able to use these methods. If you have done your homework, then these approaches can make a huge difference.
Hi J,
I understood the first question.I have a question regarding an alternate method.
If we simply substitute the value of x i.e -(1/5) in the equation and see,
we get (4/5)*(26/25)*(626/625)….with all values being “very very” slightly greater than 1.
on visualising this we see that 0.8 multiplied by 1.0x (that too decreasing) will just slightly increase 0.8 which leaves only the last option i.e 0.833.
Is this a feasible method?
If you are confident in the pattern, yes. The danger is that almost all the options are greater than 0.8. I also tried that first, but was not confident that it would stop below 1 (have made embarrassing estimation mistakes in the past 🙂 so I choose not to take this type of risk, usually), hence looked for another way!
CAT-holics 
Hey J,
In the first question, I did not understand how multiplying f(x) with (1-x) would give you 1-x^(inf)
f(x) would become (1-x)(1+x)(1+x^2)(1+x^3)….=(1-x^2)(1+x^2)(1+x^3)(1+x^4)…..=(1+x^3)(1-x^4)(1+x^4)…..
So, although the terms with even powers are taken care of, what about those with odd powers? (1+x^3)(1+x^5)(1+x^7)….
Oops yes, that is a typo. I will correct it tomorrow…thanks.
regards
J
sir, in last question how is x less that 0.5 and greater that 0.44….i already read the concept in earlier blog but there also i was having the same confusion…
1/ (2+something) 4/9
If you are not comfortable with calculation (tables, basic percentage-ratio conversion, squares, and so on), then you will NOT be able to use these methods. If you have done your homework, then these approaches can make a huge difference.
regards
J
Hi J,
I understood the first question.I have a question regarding an alternate method.
If we simply substitute the value of x i.e -(1/5) in the equation and see,
we get (4/5)*(26/25)*(626/625)….with all values being “very very” slightly greater than 1.
on visualising this we see that 0.8 multiplied by 1.0x (that too decreasing) will just slightly increase 0.8 which leaves only the last option i.e 0.833.
Is this a feasible method?
If you are confident in the pattern, yes. The danger is that almost all the options are greater than 0.8. I also tried that first, but was not confident that it would stop below 1 (have made embarrassing estimation mistakes in the past 🙂 so I choose not to take this type of risk, usually), hence looked for another way!
regards
J