99th term in a series?

Question

1/3 + 1/15 + 1/35 + 1/63 +......up to 99 terms

how do i calculate the 99 th term ?

I am not interested about the sum but interested about how to calculate the 99 th term ?

can anybody tell me whats the 99th term with explanation

Answer 1

Look at this:

Term....Value
1 . . . . . . 1/(1*3)
2. . . . . . . 1/(3*5)
3. . . . . . .1/(5*7)

So each the nth term is 1/(2n-1)(2n+1)

So the 99th term would be 1/(197)(199) = 1/39203

Answer 2

let's find a recurrence relation

3=1*3
15 = 3*5
35=5*7
63=7*9


1 is an odd number...we may write it as 2i+1 for i=0
3 may be written as 2i+3 for i=0

now ...for i=1
2i+1=3
2i+3=5

for i=2
2i+1=5
2i+3=7

so

the series is

sum (sum symbol) of 1/[(2i+1)*(2i+3)], i from 0 to n

for i=0 -> 1st term
for i=1->2nd term

so

for i=98->99th term

the 99th term is 1/[(2*98+1)*(2*98+3)]

Answer 3

99th term in a series?
1/3 + 1/15 + 1/35 + 1/63 +......up to 99 terms
=1/1.3=1/3.5+1/5.7+1/7,9................
=1/1.3=1/3,5+1/5,7+1/7.9.........
**th term of1,3,5,7.......=1+98*2=197.
99th term of 3,5,7,9 ........=3+98*2=199
required 99th term=1/197*199answer

Answer 4

bhoo