1very very good ques but no one here is so intelligent to give the ans
(1) Find value of
\left[\frac{2010!}{2009!+2008!+2007!..............3!+2!+1!} \right] where \left[. \right]
G.I.F
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5 Answers
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49I would love if Mr. "Intelligent" Pirtish added some of his intelligence to this otherwise "dumb" site!
Awaiting your reply! :P
21My hunch is that for all n>3 (integers) the following inequality is valid
(n-2)<\frac{n!}{1!+2!+...+(n-1)!}<(n-1)
We will use Principle of Mathematical Induction to prove that.
P(4) is true.
Let P(n) be true. We will prove P(n+1) is true.
We have,
(n-2)<\frac{n!}{1!+2!+...+(n-1)!}<(n-1)
\Leftrightarrow \frac{1}{n-2}>\frac{1!+2!+...+(n-1)!}{n!}>\frac{1}{n-1}
\Leftrightarrow \frac{n!(n-1)}{(n+1)!(n-2)}>\frac{1!+2!+3!+..n!}{(n+1)!}>\frac{n!n}{(n+1)!(n-1)}
\Leftrightarrow \frac{(n+1)(n-2)}{n-1}<\frac{(n+1)!}{1!+2!+3!+..n!}<n-\frac{1}{n}
Further we have n-\frac{1}{n}<n
and (n-1)<\frac{(n+1)(n-2)}{(n-1)} for all n>3
So we get (n-1)<\frac{(n+1)!}{1!+2!+3!+...+n!}<n
P(n+1) is true. So, statement is proved.
Therefore \left[\frac{n!}{1!+2!+3!+..(n-1)!} \right] = (n-2) for all n>3
So, we get the required answer as 2008
