Evaluate
3/(1! +2! +3!) + 4/(2! +3! +4!) +
5/(3! +4! +5!) + ... + 2001/(1991! +2000! +2001!).
To evaluate the given expression, we need to first understand the factorial (!) notation. In mathematics, the factorial of a non-negative integer n, denoted as n!, is the product of all positive integers less than or equal to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.
Now, let's simplify the expression step by step. Notice that in each term, the denominator consists of three consecutive factorials, and the numerator is an integer that is one greater than the largest factorial in the denominator.
First, let's evaluate the expression for the first term:
3/(1! + 2! + 3!) = 3/(1 + 2 + 6) = 3/9 = 1/3.
Next, let's examine the second term:
4/(2! + 3! + 4!) = 4/(2 + 6 + 24) = 4/32 = 1/8.
Following the same pattern, we can observe that each subsequent term can be simplified in the same manner. Let's simplify one more term:
5/(3! + 4! + 5!) = 5/(6 + 24 + 120) = 5/150 = 1/30.
Based on these calculations, we can identify a pattern: each term consists of the number 1 divided by a multiple of 3. The numerator increases by 1 in each subsequent term, while the denominator increases by a larger value due to the multiplying effect of the factorials.
To evaluate the entire expression efficiently, we can write it in a general form. Let k represent any given term, where k ranges from 1 to 2001. The k-th term can be expressed as:
(k + 2)/(k! + (k + 1)! + (k + 2)!)
Now, we can focus on the denominator:
k! + (k + 1)! + (k + 2)!
= (k!) + [(k + 1)(k!)] + [(k + 2)(k + 1)(k!)]
= k! + (k + 1)(k!) + (k + 2)(k + 1)(k!)
= k![1 + (k + 1) + (k + 2)(k + 1)]
= k!(k + 3)(k + 2)(k + 1)
Replacing the denominator in the k-th term:
(k + 2)/(k!(k + 3)(k + 2)(k + 1))
Notice that the numerator can be expressed as (k + 2) = (k + 3) - 1.
Substituting this back into the expression:
((k + 3) - 1)/(k!(k + 3)(k + 2)(k + 1))
= (k + 3)/(k!(k + 3)(k + 2)(k + 1)) - 1/(k!(k + 3)(k + 2)(k + 1))
Now, we can cancel out the (k + 3) terms in both the numerator and denominator:
1/k!(k + 2)(k + 1) - 1/(k!(k + 2)(k + 1))
The denominators in both terms are the same, so we can combine them into a single fraction:
[1 - 1/(k + 1)]/(k!(k + 2)(k + 1))
We can further simplify the expression in the square brackets:
[(k + 1)/(k + 1) - 1/(k + 1)]/(k!(k + 2)(k + 1))
= (k + 1 - 1)/(k + 1)/(k!(k + 2)(k + 1))
= k/(k + 2)/(k!(k + 2)(k + 1))
= 1/(k + 2)/(k!)
Now, let's rewrite the entire expression using this simplified form:
1/3 + 1/8 + 1/30 + ... + 1/(k + 2)/(k!) + ... + 1/(2001 + 2)/(2001!)
The common denominator for each term is the product of all the factorials, (k!) in this case, so we can express the entire expression over this single denominator:
(1/(3(k!)) + 1/(8(k!)) + 1/(30(k!)) + ... + 1/((k + 2)(k!)) + ... + 1/((2001 + 2)(2001!)))
Now, we can rewrite the expression using summation notation:
∑[1/((k + 2)(k!))] for k = 1 to 2001
To evaluate this summation formula, we can use a computer program or a mathematical software package like MATLAB or Python.
Alternatively, if you need an approximate value, you can plug in the upper limit (k = 2001) into the formula and calculate the result. Keep in mind that the resulting value might be very large, so avoid calculating it manually, as this would be tedious and prone to errors.