62This one is a good question..
someone give the whole explanation ...
24At first look i too wrote answer to be Zero....but its impossible to tell if (n!)2-n! is a perfect square when n→∞
and answer given is 1
here is my explaination..tell if its right or not
sin2(Ï€√(n!)2-n!)
sin2(Ï€.n!√1-1n!)
sin2(Ï€.n!(1-1n!)1/2)
sin2(Ï€.n!(1-12.n!+....))
sin2(Ï€(n!-12+...))
limn→∞sin2(π(n!-12+...))
=> limn→∞sin2(π.n!-π2+...)
Now n! will be an integer even when n→∞ and all terms greater than π2 become zero as n→∞
So we can write it as sin2(kπ-π2)
=> sin2(-Ï€/2)
=>1
62
Lokesh Verma
·2009-12-27 21:09:48
i somehow dont tend to agree to the answer
because the term under the squareroot needs to be a perfect square..
so (n!)2-n! = n!(n!-1)
now this cannot be a perfect square bcos
t(t-1) is never a perfect square for any integer greater than 1! (Why?)
Hence the answer is zero.. for the question..
24
eureka123
·2009-12-28 08:29:04
sir u r contradiciting urself ....
on one side u said t(t-1) can never be perfect square....and on other side u say answer to be zero...
tell me if t(t-1) is never perfect square => √t(t-1) is never integer => angle will never be integer multiple of pi..so answer will never be zero
341\sin^2 (\pi \sqrt{(n!)^2-n!}) = \sin^2 (-\pi \sqrt{(n!)^2-n!}) = \sin^2 (n! \pi - \pi \sqrt{(n!)^2-n!})
= \sin^2 \frac {n!\pi}{n! + \sqrt{(n!)^2-n!}}
= \sin^2 \frac {\pi}{1 + \sqrt{\left(1-\frac{1}{n! \right)}}} \rightarrow \sin^2 \frac{\pi}{2} = 1
