Final answer:
The sum of the given series 1/(1x3) + 1/(2x4) + 1/(3x5) + .... + 1/(98x100) is approximately 0.3232.
Step-by-step explanation:
The given series is 1/(1x3) + 1/(2x4) + 1/(3x5) + .... + 1/(98x100). To find the sum of this series, we need to first determine a pattern by looking at the terms. We can observe that the denominators of the fractions are increasing by 2, starting from 3. Therefore, the nth term of the series can be written as 1/(n x (n+2)).
Now, let's find the sum:
S = 1/(1x3) + 1/(2x4) + 1/(3x5) + .... + 1/(98x100)
S = 1/[(1x2) x (2x3)] + 1/[(2x3) x (3x4)] + 1/[(3x4) x (4x5)] + .... + 1/[(98x99) x (99x100)]
Simplifying, we get:
S = 1/(2x3) + 1/(3x4) + 1/(4x5) + .... + 1/(98x99)
Now, if we subtract this series from the original series, we can cancel out terms and simplify:
S - S = 1/[1x3 - 2x3] + 1/[2x4 - 3x4] + 1/[3x5 - 4x5] + .... + 1/[98x100 - 99x100]
Which gives:
0 = 1/[1x3] - 1/[99x100]
Therefore, the given series is a telescoping series and all terms cancel out, except for the first and last term. The sum of the series is equal to the difference between the first and last term:
S = 1/[1x3] - 1/[99x100]
Calculating the values:
S = 1/3 - 1/[99x100]
So, the sum of the series is approximately 0.3333 - 0.0101 = 0.3232.