Find the sum:
[tex]$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\ldots+\frac{1}{110 \times 111}=\frac{m}{n}$[/tex]
Where [tex]$\operatorname{gcd}(m, n)=1$[/tex]
What is [tex]$m+n$[/tex]?
A) 221
B) 207
C) 244
D) 229
E) 208
Answer (2)
The sum evaluates to \frac{110}{111}, with m=110 and n=111. Thus, m+n equals 221. Therefore, the chosen answer is option A) 221.
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Decompose each fraction in the sum into partial fractions: k ( k + 1 ) 1 = k 1 − k + 1 1 .
Recognize the sum as a telescoping sum where intermediate terms cancel out.
Simplify the sum to 1 − 111 1 = 111 110 , so m = 110 and n = 111 .
Calculate m + n = 110 + 111 = 221 .
Explanation
Problem Analysis We are given the sum of fractions: 1 × 2 1 + 2 × 3 1 + … + 110 × 111 1 = n m where g c d ( m , n ) = 1 . Our goal is to find the value of m + n .
Partial Fraction Decomposition Each term in the sum can be written as a difference of two fractions using partial fraction decomposition: k ( k + 1 ) 1 = k 1 − k + 1 1 This allows us to rewrite the sum as: k = 1 ∑ 110 k ( k + 1 ) 1 = k = 1 ∑ 110 ( k 1 − k + 1 1 )
Telescoping Sum The sum is a telescoping sum, meaning consecutive terms cancel each other out. Writing out the first few and last few terms, we have: ( 1 1 − 2 1 ) + ( 2 1 − 3 1 ) + … + ( 110 1 − 111 1 ) Notice that − 2 1 cancels with + 2 1 , − 3 1 cancels with the next + 3 1 , and so on, until − 110 1 cancels with + 110 1 .
Simplifying the Sum After cancellation, the sum simplifies to: 1 − 111 1 Calculating this value, we get: 1 − 111 1 = 111 111 − 1 = 111 110
Finding m+n Since g c d ( 110 , 111 ) = 1 , the fraction 111 110 is in its simplest form. Therefore, m = 110 and n = 111 .
Finally, we compute m + n :
m + n = 110 + 111 = 221
Final Answer Thus, the value of m + n is 221.
Examples
Telescoping sums are useful in many areas of math and physics. For example, imagine you are calculating the total displacement of an object that moves in discrete steps. If the displacement at each step can be expressed as a difference, the total displacement simplifies to the difference between the initial and final positions, much like our telescoping sum. This principle is also used in simplifying complex series in calculus and physics problems.
