Using a pattern, what is the sum of:
1/1 + 1/2 x 1/3 +1/3 x 1/4 + ....... + 1/99 x 1/100 + 1/100 x 1/101?
We can find a very good approximation of this sum by using the formula:
Hn≈(2n+1)[arccot(2n+1)]+(1/2)[ln(n2+n+1/2)]+(γ−1)
For n=100 , we take:
H100≈5.18737751…
The exact value of this sum is the given by the fraction:
144666362795203511602215180431041314477112788815009188499086581352357412492142272
This fraction has the estimated value of 5.18737751763962…
The integer part of it is 5 .
mmmhhhh?
I see a telescopic series
1/1 + 1/2 x 1/3 +1/3 x 1/4 + ....... + 1/99 x 1/100 + 1/100 x 1/101
= 1 + (1/2)(1/3) + (1/3)(1/4) + ... + (1/99)(1/100) + (1/100)(1/101)
the general term is 1/(n(n+1))
using partial fractions I can break this up as 1/n - 1/(n+1)
so we have
1 + Sigma [1/n - 1/(n+1) ] for n = 2 to 100
= 1 + (1/2-1/3) + (1/3-1/4) + (1/4-1/5) + ... + (1/99-1/100) + (1/100-1/101)
= 1 + 1/2 - 1/101 , all the other terms in between drop out
= 301/202 or appr. 1.49
( I have no idea where "helper" got his stuff from, just adding the first 4 terms
by calculator gives us 1.3 )
okay thx mathhelper
To find the sum of the given pattern, let's analyze the terms:
1/1 + 1/2 x 1/3 + 1/3 x 1/4 + ....... + 1/99 x 1/100 + 1/100 x 1/101
From the pattern, we can observe that each term consists of two fractions multiplied together. The first fraction in each term has a numerator of 1 and the denominator increases by 1 with each subsequent term. The second fraction in each term also has a numerator of 1 and the denominator increases by 1 compared to the first fraction's denominator.
Let's rewrite the given pattern separately to see if we can find any patterns:
1/1 + (1/2) x (1/3) + (1/3) x (1/4) + ...... + (1/99) x (1/100) + (1/100) x (1/101)
Looking at the pattern, we can see that each term can be written in the form of:
(1/n) x (1/(n+1))
The denominator of the first fraction is n, and the denominator of the second fraction is (n + 1).
We can simplify this further:
(1/n) x (1/(n + 1)) = 1 / (n(n + 1))
Now, let's rewrite the entire pattern in this simplified form:
1/(1 x 2) + 1/(2 x 3) + 1/(3 x 4) + ....... + 1/(99 x 100) + 1/(100 x 101)
Notice that each term has the same format: 1 divided by the product of two consecutive numbers.
To simplify the sum of these terms, we can use the concept of telescoping series. In telescoping series, many terms cancel out each other when simplified.
Let's take the first few terms as an example:
1/2 - 1/3 + 1/3 - 1/4 + ...... + 1/100 - 1/101
In this sequence, notice that each term cancels out the previous term except for the first term (1/2) and the last term (-1/101).
Hence, the sum of the pattern simplifies to:
1/2 - 1/101
Now we can calculate the final answer:
1/2 - 1/101 = (101 - 2) / (2 x 101) = 99 / 202
Therefore, the sum of the given pattern is 99/202.