Induction (1 Viewer)

[dawso]

ok, got this "harder 3unit" induction question, without boring peoples with all of it, ill get 2 the point....

i am tryin 2 prove: 1 + 1/sqrt2 + 1/sqrt3 + ... + 1/sqrt(n) > 2 [ sqrt (n+1) - 1]

in gettin that it is true for n=1, is it acceptable to just put tleft hand side and right hand side approximations by calculator....or justthat it is obviously more, or do u NEED to take it 2 one side and get that it is greater than 0...

its not hard, but coming from a question 8....may save much needed time if it aint needed...

(for those of u with it, its taken from 1985 hsc, and is in the inequalities notes from TSD)

cheers...

 

[Templar]

In this case, calculator should be fine.

 

[shafqat]

For n = 1, we're trying to prove 1 > 2(sqrt2 - 1)
That is 3 > 2sqrt2, which is true because 9 > 8

 

[Slidey]

Or the method he used is:

LHS=1, RHS=2(sqrt2 - 1)
LHS-RHS=3-2sqrt2. Unrationalise:
(3-2sqrt2)(3+2sqrt2)/(3+2sqrt2)
=1/(3+2sqrt2)>0
.'. LHS > RHS

 

[dawso]

lol, thats not what "u used" its just what mahoney says, but anyway....i get that, i was just wonderin if its necessary.... i just finished all the examples in those notes then

 

[dawso]

ok, ive got another induction thats killing me....

from 1990 hsc....

use MI to show that:

sqrt 1 + sqrt 2 + ... + sqrt (n) <= [(4n + 3) sqrt (n)] / 6

cheers

 

[shafqat]

.....................

 

[dawso]

yeah, ive got that, lol, i just dont get what they have done so maybe someone with more steps??

they simply put in the k+1 with assumption and then jump 2 what they want? i cant get the in between steps out

 

[Antwan23q]

wtf is with those pictures. anyone mods care to deleat them?

 

[dawso]

yeah, theyve banned him....shud get around 2 deleting it soon