How do you evaluate the sum represented by #sum_(n=1)^5n/(2n+1)# ?

Question

5021 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-13T21:32:46

First expand the series for each value of n #n/(2n+1)#=#1/(2(1)+1#+#2/(2(2)+1#+#3/(2(3)+1#+#4/(2(4)+1#+#5/(2(5)+1#

Next, perform the operations in the denominator...

#1/(3)#+#2/5#+#3/7#+#4/9#+#5/11#

Now, to add fractions we need a common denominator... in this case it's #3465#

Next, we have to multiply each numerator and denominator by the missing components...

#1/3# gets multiplied by #1155# giving #1155/3465#

(Divide the #3465# by #3# to get #1155# and divide the rest by the given denominator.)

#2/5*693/693=1386/3465, 3/7*495/495=1485/3465, 4/9*385/385=1540/3465 and 5/11*315/315=1575/3465#

Now simply add the numerators together... #(1155+1386+1485+1540+1575)/3465#

giving #7141/3465#.