Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

In this problem we evaluate the series:
1/(1x3)+1/(2x4)+1/(3x5)...+1/(98x100)

A. Notice that each fraction in the sum has the form 1/(n(n+2)) for some positive integer n. Find constants A and B such that 1/(n(n=2)=A//n+B/(n+2)

B. Use your answer to part A to find the desired sum

Sagot :

sqdancefan

Answer:

  A.  1/(n(n+2)) = (1/2)/n - (1/2)/(n+2)

  B.  14651/19800

Step-by-step explanation:

A.

The coefficients of the partial-fraction expansion can be found from ...

  f(n) = 1/(n(n+2)) = A/n +B/(n+2)

  n·f(x) = 1/(n+2) = A +Bn/(n+2)

For n=0, this becomes ...

  1/(0 +2) = A = 1/2

__

Similarly, ...

  (n+2)·f(n) = 1/n = A(n+2)/n +B

For n = -2, this becomes ...

  1/(-2) = B = -1/2

The n-th term can be written ...

  1/(n(n+2)) = (1/2)/n - (1/2)/(n+2)

__

B.

The sum is ...

  1/(1·3) +1/(2·4) +1/(3·5) +... +1/(98·100)

  = 1/2(1/1 -1/3 +1/2 -1/4 +1/3 -1/5 +... +1/98 -1/100)

  = 1/2((1/1 +1/2 +1/3 +...1/98) -(1/3 +1/4 +1/5 +...+1/100)

We notice that terms 3..98 cancel, so the sum is ...

  = 1/2(1/1 +1/2 -1/99 -1/100) = (1/2)(3/2 -199/9900)

  = 14651/19800

View image sqdancefan
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.