IIT JEE , AIEEE Forum - Mathematics , Algebra - solve if you can shortest method appreciated

ziauddin75

2^1/4 * 4^1/8 * 8^1/16 upto infinity

and

min. value of x^4+1/x^2 is what

sandeepramesh

for the first one, basically its 2^{1/4+2/8+3/16..............) which is 2^{infinite AGP} :)

sandeepramesh

by summing it up you get 2^4 = 16 :)

Is it right? For further details see AGP

ziauddin75

answer full

ziauddin75

wring one

sandeepramesh

first tell me is 16 right? and i have already told you its an AGP and have given you the link for i too :)

sandeepramesh

im very sorry but the answer is 2. I substituted the values wrongly. Dont tell me 2 is also wrong

ziauddin75

what abt second one

try this one too

check whether 111111111111----------(91 times)
prime hai ki nahin

sandeepramesh

its not a prime simply bcos 1111111 (7 1's) and 1111111111111 (13 1's) clearly divide it :)

EDIT: is 2 right btw?

studyid

yeah sandeep 2 is the right ans.

ziauddin75

how can you say its clearly divisable by 31 ant trick share it then plzzzzz

sandeepramesh

isnt it obvious that 1111111 and 1111111111111 divide

1111.........1 (91 1's) ?????

ziauddin75

theek kaha aisa kuch socha nahin the pehle hi ques. tough samagh liya tha