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one can show that the relation between first Chern class and second Chern class of $CP^n$ is

$\frac{2(n+1)}{n} c_2 (M)=c_1 (M)^2$

here $c_1 (M)^2=c_1 (M)∧c_1 (M)$. So is there any recurrence formula for i-th Chern class of $CP^n$ ?

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There is no need for a recurrence formula. If $h\in \mathrm H^2(\mathbb C\mathbb P^n, \mathbb Z)$ is the Poincaré dual of a hyperplane section, then $h^i$ generates $\mathrm H^{2i}(\mathbb C\mathbb P^n, \mathbb Z)$ for all $i = 0, \dots, n$. Then it follows from the Euler sequence that $$ \mathrm c_i(\mathbb C\mathbb P^n) = \binom{n+1}{i}h^i. $$

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    $\begingroup$ Angelo -- I completely agree with what you say. However, perhaps the OP was also thinking about how to express powers of $c_1$ in terms of the other classes: a problem that leads directly to the Segre classes (in the relative setting). $\endgroup$
    – Jason Starr
    Commented Jul 1, 2012 at 14:57
  • $\begingroup$ Dear Jason, would you please explain with more details $\endgroup$
    – user21574
    Commented Jul 1, 2012 at 17:21
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    $\begingroup$ To Jason: well, in this particular case $c_1^i$ is a multiple of $c_i$, so you don't see the Segre classes. $\endgroup$
    – Angelo
    Commented Jul 1, 2012 at 17:30
  • $\begingroup$ Dear Jason, for more information you can see the reason of Angelo in following explanatory paper math.columbia.edu/~tosatti/cpn.pdf $\endgroup$
    – user21574
    Commented Jul 1, 2012 at 18:11
  • $\begingroup$ Angelo: Of course you are right in the absolute setting, I was guessing (probably wrongly) that the OP had in mind something broader (the relative case, which leads to Segre classes, as I am sure you already know). $\endgroup$
    – Jason Starr
    Commented Jul 2, 2012 at 13:53

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