This one is relatively simpler compared to the other question we have solved..
\left[\frac{n}{3}\right]+\left[\frac{n+2}{6}\right]+\left[\frac{n+4}{6}\right]=\left[\frac{n}{2}\right]+\left[\frac{n+3}{6} \right]
1LHS \ :\ \left[\frac{n}{3} \right]+\left[\frac{n}{6}+\frac{1}{3}\right]+\left[\frac{n}{6}+\frac{2}{3}\right]=\left[\frac{n}{3} \right]+\left[\frac{n}{2} \right]-\left[\frac{n}{6} \right]
R.H.S. \ :\ \left[\frac{n}{2} \right]+\left[\frac{n}{6}+\frac{1}{2} \right]=\left[\frac{n}{2} \right]+ \left[\frac{n}{3} \right]-\left[\frac{n}{6}\right]
used this one :
http://targetiit.com/iit-jee-forum/posts/10-09-09-greatest-integer-function-11066.html
62I am giving a "pink slip" but the fact is that it is much simpler...
I mean this one can be solved by simply taking n between "" and "' and then showing that it holds for the rest..
Use the idea that prophet sir gave in that question..
1okie.
Let\ F(n)=\left[\frac{n}{3}\right]+\left[\frac{n+2}{6}\right]+\left[\frac{n+4}{6}\right]-\left[\frac{n}{2}\right]-\left[\frac{n+3}{6} \right]
Now\ we\ note\ that\ F(n)=F(n+6).
1Now it is ok.
thanks for the correction.
Actually I copied the url from nishant bhaiyya's post and edited it directly, so missed out the other sign.
3sir i din see the previous discussion,on seeing the question i think it can be proved if we take two cases if n is a multiple of 6 and it is not.so we can do the rest part of the soln,though it is not elegant method of solving,i felt this is the easiest soln.
