66
kaymant
·2009-08-17 04:32:11
indeed it is:) justification?
341
Hari Shankar
·2009-08-17 04:50:50
Ok, i worked with this reckoning that as x becomes large, each of the factors are almost equal so that \sqrt [n] {(x+2)(x+4)...(x+2n)} will tend to the AM of the factors which is x+n+1.
Likewise, the second term tends to x+n+1/2
Hence their difference is 1/2
Now, between the earlier post and this I arrived at the proof of
\sqrt [n] {(x+a_1)(x+a_2)...(x+a_n)} \rightarrow x + A as x \rightarrow \infty where
A = \frac{a_1+a_2+...+a_n}{n}
This can easily be attempted by the students. So, letz wait
66
kaymant
·2009-08-17 04:59:17
Yes... you are right (which you usually are [1]). And of course, if one proves that last result of yours, he/she will reach the conclusion. Lets wait.
62
Lokesh Verma
·2009-08-17 05:19:42
btw this was proved for n=4 sometime back :)
11
Devil
·2009-08-17 10:23:07
{(x+a1)(x+a2)...(x+an)}1/n = x+A, as x →∞, if the proof of this is reqd, it's damn easy....
Consider f(x)=R.H.SL.H.S, now dividing N & D by x we have our result,
But is this the Qsn at all...or is to find the limit as x→∞ of {(x+a1)(x+a2)...(x+an)}1/n ???
1
rahul1993 Duggal
·2009-08-27 21:21:36
here is my sir's solution (you can download the file)
http://www.mediafire.com/?ge3zu7zt2yh