sum of series?

s = 1/(1*3) + 1/(3*5) + 1/(5*7) + 1/(7*9)+ ... + 1(197*199)

Sum of one term: 1/3

Sum of two terms: 2/5

Sum of three terms: 3/7

Sum of four terms: 4/9

Sum of five terms: 5/11

etc.

By observation:

Sum of n terms: n/(2n-1)

Therefore:

Sum of 99 terms: 99/199

Correct answer is 1) 99/199

Edit: It appears my answer is very similar to Babatunde... but we came up with them independently. I think mine is just a little clearer however.

S = SUM

S1 = 1/3

S2 = 1/3 + 1/15 = 2/5

S3 = 1/3 + 1/15 + 1/35 = 3/7

S4 = 1/3 + 1/15 + 1/35 + 1/63 = 4/9

SK = k/2k+1

Therefore...

S99 = 99/2*99+1 = 99/198+1 = 99/199

#1

The series can be written as

1/2[1-1/3]+1/2[1/3-1/5]+1/2[1/5-1/7]+...+1/2[1/197-1/199]

Here we see that the 2nd fraction of each term is cancelled with the first fraction of the next term...So what remains ultimately is1/2[1-1/199]=99/199

So the correct answer is 1

Answer is 1) 99/199

Clearly the ganeric term is :

T(n) = 1/ (4 n^2 -1) = 1/ ((2n-1)(2n+1))

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

thus T(1) = 1/2( 1- 1/3)

T(2) = 1/2( 1/3 - 1/5)

.

.

.

T(n)= 1/2 (1/(2n-1) - 1/(2n+1))

Summing up.. we find that the middle terms cancel out :)

thus

Sum = 1/2( 1- 1/(2n+1))

Thus for n =99

Sum = 1/2( 1 - 1/199) = 99/199 :)

babatunde is correct, but another way to do it is to split it up using partial fractions & evaluate the individual sums.

your series can be represented as:

SUM(1/(2n-1)*(2n+1)) from n=1 to n=99

the nth term is the expression above, i.e.

1/(2n-1)*(2n+1) = A/(2n-1) + B/(2n+1) (using partial fractions)

by doing some mathematical work, we can show that A=(1/2), B=(-1/2). Hence, our nth term becomes:

(1/2)*(1/(2n-1) - 1/(2n+1))

Hence, our series can be rewritten as

SUM((1/2)*(1/(2n-1) - 1/(2n+1))) from n=1 to n=99

or,

(1/2)*(SUM(1/(2n-1)) - SUM(1/(2n+1)))

expanding the series, we have:

(1/2)*( (1 + 1/3 + 1/5 ... 1/195 + 1/197) - (1/3 + 1/5 ... 1/197 + 1/199) )

If you observe the expression, you can easily cancel out many terms like 1/3, 1/5, 1/195, 1/197 etc. leaving you with:

(1/2)*(1 - 1/199)

= 198/(199*2)

= 99/199

Hence, it is choice (1)

You can write the sum as

1/[(2n+1)(2n-1)]

The series starts at n = 1.

The sum from n = 1 to n = 99 is

99/199

1/3+1/15+1/35+.....=1/[(2^2)-1] + 1/[(4^2)-1]+ 1/[(6^2)-1]..

this seri is a second rooth equations. and also the denominator is getting bigger(that means zero) so that seri can not determinate.

I would "vote" for answer 1)

-----------------------------

i was lucky to eliminate the other 3 grossly wrong answer in order to be the 1st with correct one :))

let me comment on the rest -- clearly there 2 type of solutions:

- by elimination of intermediate terms and

- by finding formula for the sum

logically the first method is superior, why? - because the second methods needs one more step: you have to prove (not just observe) this formula using the method of complete mathematical induction

Mate that Babutunde guys solution is awesome thats the safest and best way to solve seri...

4) cannot be determined