91Ó°ÊÓ

Given that$P(X=n)=\frac{4}{n(n+1)(n+2)},nâ‰\yen 1$.

$\underset{n=1}{\overset{+∞}{∑}}P(X=n)$

$=\underset{n=1}{\overset{+∞}{∑}}\frac{4}{n(n+1)(n+2)}$

$=\underset{n=1}{\overset{+∞}{∑}}\left(\frac{4}{n(n+1)}-\frac{4}{n(n+2)}\right)$

$=\underset{n=1}{\overset{+∞}{∑}}\left(\frac{4}{n}-\frac{4}{n+1}-\frac{4}{n(n+2)}\right)$

$=\underset{n=1}{\overset{+∞}{∑}}\left(\frac{4}{n}-\frac{4}{n+1}-\frac{2}{n}+\frac{2}{n+2}\right)$

$=2+1+2\underset{n=3}{\overset{+∞}{∑}}\frac{1}{n}-2-4\underset{n=3}{\overset{+∞}{∑}}\frac{1}{n}+2\underset{n=3}{\overset{+∞}{∑}}\frac{1}{n}$

$=1$

The answer is $\underset{n=1}{\overset{+∞}{∑}}P\left(X=n\right)=$$1$

Given that $P(X=n)=\frac{4}{n(n+1)(n+2)},nâ‰\yen 1$

The expected value is the sum of each possibility n, with its possibility .

$E\left(X\right)=\underset{i=1}{\overset{+∞}{∑}}nP(x=n)$

$=\underset{n=1}{\overset{+∞}{∑}}\frac{4n}{n(n+1)(n+2)}$

$=\underset{n=1}{\overset{+∞}{∑}}\frac{4}{(n+1)(n+2)}$

$=4\underset{n=2}{\overset{+∞}{∑}}\frac{1}{n}-4\underset{n=3}{\overset{+∞}{∑}}\frac{1}{n}$

= $4×\frac{1}{2}$

$=2$

The answer is$E\left(X\right)=2$

Given that$P(X=n)=\frac{4}{n(n+1)(n+2)},nâ‰\yen 1$

$E\left(X^2\right)=\underset{i=1}{\overset{+∞}{∑}}{n}^{2}P(x=n)$

$=\underset{n=1}{\overset{+∞}{∑}}\frac{4n}{n(n+1)(n+2)}$

$=\underset{n=1}{\overset{+∞}{∑}}\frac{4n}{(n+1)(n+2)}$

$=2+4\underset{n=3}{∑}\frac{1}{n}$

$=2+4(+∞)$

=$+∞$

The answer is$E\left(X^2\right)=$$+∞$