$\sum\limits^\infty_{n=3}\frac{6}{n^2-1}=\sum\limits^\infty_{n=3}\frac{6}{(n-1)(n+1)}\\\\=\sum\limits^\infty_{n=3}\frac{6}{2}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=\frac{6}{2}\sum\limits^\infty_{n=3}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=3[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+...]\\\\=3[\frac{1}{2}+\frac{1}{3}]=3[\frac{3+2}{6}]=3[\frac{5}{6}]\\\\=\frac{15}{6}=\frac{5}{3}$

5 0

3 years ago

Rewrite the statement in mathematical notation. (Let n be the number of cases of Bangkok flu and t be time.) There are presently

Darya [45]

Let us say that t is number of months and n is the number of cases of Bangkok flu, therefore the equation we can create here is:

dn / dt = 10 t + 450

<span>where 10 represents the slope and 450 represents the y-intercept</span>

7 0

3 years ago

In ΔABC, a = 300 inches, ∠B=45° and ∠C=41°. Find the length of c, to the nearest 10th of an inch.

dezoksy [38]

Answer:

197.3

Step-by-step explanation:

4 0

3 years ago