1/1*2 + 1/2*3 + 1/3*4 + . . . + 1/2000*2001 + 1/2001*2002
A. 1/2002
B. 1999/2002
C. 2001/2002
D. 1
I can do this, I want to know how to do it a faster way... thanks!!
How are you doing it?
Perhaps you are doing it the fastest way.
What is your choice of answers?
I just add all of them, but it is taking a long time!
ok, that is not a good idea.
How about this....
let Sum(n) denote the sum of n terms,
sum(1) = 1/2
sum(2) = 1/2 + 1/6 = 4/6 = 2/3
sum(3) = 2/3 + 1/12 = 9/12 = 3/4
sum(4) = 3/4 + 1/20 = 16/20 = 4/5
notice any pattern yet ???
let's do one more
sum(5) = 4/5 + 1/30 = 25/30 = 5/6
I am very confident that
sum(n) = n/(n+1)
and since we see that there are 2001 terms
sum(2001) = 2001/2002
Oh I get it, thanks!
To find a faster way to solve this problem, we can observe that each term in the series has a common structure. Let's break it down step by step:
1/1 * 2 + 1/2 * 3 + 1/3 * 4 + ... + 1/2000 * 2001 + 1/2001 * 2002
If we look closely, we can see that the nth term is given by:
1/n * (n+1)
Now, we can simplify this further by canceling out common factors:
1/(n * n+1)
At this point, we see that each term is the reciprocal of the product of two consecutive numbers, i.e., n and n+1.
Now, let's express the given series in terms of this simplified expression:
1/(1 * 2) + 1/(2 * 3) + 1/(3 * 4) + ... + 1/(2000 * 2001) + 1/(2001 * 2002)
Now, we notice that some terms can be combined:
(1/(1 * 2) + 1/(2 * 3)) + (1/(3 * 4) + 1/(4 * 5)) + ... + (1/(2000 * 2001) + 1/(2001 * 2002))
We can group terms in pairs and observe that a pattern is emerging:
[(1/(1 * 2) + 1/(2 * 3)) + (1/(3 * 4) + 1/(4 * 5))] + ... + [(1/(1999 * 2000) + 1/(2000 * 2001)) + (1/(2001 * 2002))]
Notice that each pair of terms inside the parentheses cancels out certain factors:
[(1/2) + (1/3)] + [(1/4) + (1/5)] + ... + [(1/2000) + (1/2001)] + [(1/2001 * 2002)]
We can continue this pattern until we reach the final pair:
(1/2000 + 1/2001) + (1/2001 * 2002)
Now, we can simplify each pair's expression:
(2001/2) + (2002/3) + ... + (1/2001 * 2002)
To calculate this sum, we can factor out the common term of 2002:
2002 * [(1/2) + (1/3) + ... + (1/2001)]
Now, we have a much simpler expression:
2002 * (1/2 + 1/3 + ... + 1/2001)
We can now evaluate this sum much faster by using the formula for the harmonic series:
Sum = n(Hn) = n * (ln(n) + γ) [approximately for large n]
Using this formula, where n is 2001:
Sum = 2001 * [ln(2001) + γ] = 2001 * (7.602 + γ) ≈ 2001 * 8.854 ≈ 17716.151
Finally, multiplying this result by 2002 gives us:
2002 * 17716.151 ≈ 35488373.502
Therefore, the sum of the series is approximately 35488373.502.
To match this result with the given answer choices, we can simplify the approximate sum:
35488373.502 ≈ 35488373
Now, comparing this to the options:
A. 1/2002 ≈ 0.0005
B. 1999/2002 ≈ 0.9945
C. 2001/2002 ≈ 1.0005
D. 1
The closest answer to the approximate sum is option B. 1999/2002.
Therefore, the correct answer is B. 1999/2002.