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.
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]
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10 Answers
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).
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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.