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605-655 (Medium)|   Algebra|   Arithmetic|      
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Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll Updated on: 15 Dec 2023, 04:24
Originally posted by liviofrol on 12 Dec 2023, 15:09.
Last edited by Bunuel on 15 Dec 2023, 04:24, edited 1 time in total.
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Given that \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\), which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ ... +\frac{(2)(29)}{(28)(29)(30)}\)?

​(A) \(\frac{14}{15}\)

(B) \(\frac{2}{35}\)

(C) \(\frac{117}{35}\)

(D) \(\frac{232}{35}\)

(E) \(\frac{232}{969}\)

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umangshah
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Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 15 Dec 2023, 19:08
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liviofrol
Given that \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\), which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)?

​(A) \(\frac{14}{15}\)

(B) \(\frac{2}{35}\)

(C) \(\frac{117}{35}\)

(D) \(\frac{232}{35}\)

(E) \(\frac{232}{969}\)

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On expanding the equation using \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\) , we can easily find a pattern.

Term 1: \(\frac{(2)(29)}{(5)(6)(7)}\)= \(\frac{29}{(5)(6)}-\frac{29}{(6)(7)}\)
Similarly term 2: \(\frac{(2)(29)}{(6)(7)(8)}\)=\(\frac{29}{(6)(7)}-\frac{29}{(7)(8)}\)

Pattern Continues upto \(\frac{(2)(29)}{(28)(29)(30)}\)

We can see that on adding term 1 and 2 \(\frac{29}{(6)(7)}\) cancels out. The same continues leaving us with only 1st and last term to consider.

Therefore, \(\frac{29}{(5)(6)}-\frac{29}{(29)(30)}= \frac{29(29-1)}{(29)(30)}=\frac{28}{30}=\frac{14}{15}\)
Option A is correct.
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Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 21 Dec 2023, 23:53
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Bunuel do you have similar questions for practice?
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Try checking HERE.

Hope it helps.
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Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 13 Dec 2023, 06:55
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liviofrol
Which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)?

​A) \(\frac{14}{15}\)

B) \(\frac{2}{35}\)

C) \(\frac{117}{35}\)

D) \(\frac{232}{35}\)

E) \(\frac{232}{969}\)
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USE OPTIONS

Use options as they are wide spread.
\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)
\(\frac{(2)(29)}{(5)(6)(7)}~=\frac{1}{4}......\frac{(2)(29)}{(6)(7)(8)}~=\frac{1}{6}......\frac{(2)(29)}{(7)(8)(9)}~=\frac{1}{9}...... … \frac{(2)(29)}{(28)(29)(30)}~=\frac{1}{420}......\)

(1) The value has to be greater than 1/3, so discard B and E
(2) The first value is 1/3, then it reduces drastically with each successive term...1/3, 1/4, 1/9, all other terms are less than 1/10, and it reaches 1/140
Clearly, the sum will around 1 or at the max 1.5. Only possibility 14/15
(3) For the answer to be 117/35 or nearly 3 for the sum of 24 terms, each term should be nearly 3/24 or 1/8. But only 2 terms are greater than 1/8.

Approximation
Take 29 as 30 and the actual answer answer should be lesser than what you get.
\(\frac{(2)(30)}{(5)(6)(7)}+\frac{(2)(30)}{(6)(7)(8)}+\frac{(2)(30)}{(7)(8)(9)}+ … +\frac{(2)(30)}{(28)(29)(30)}\)
\(\frac{10}{(5)(7)}+\frac{10}{(7)(8)}+\frac{10}{(7)(4)(3)}+ … +\frac{10}{(14)(30)}\)
\(\frac{1}{3.5}+\frac{1}{5.6}+\frac{1}{8.4}+\frac{1}{12}+\frac{1}{16} … +\frac{1}{(140)}\)
All options can be discarded except 14/15

Proper method

\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)

Now, \(\frac{1}{(5)(6)(7)}=\frac{1}{2*5}-\frac{1}{6}+\frac{1}{2*7}............\frac{1}{(6)(7)(8)}=\frac{1}{2*6}-\frac{1}{7}+\frac{1}{2*8}............\frac{(2)(29)}{(7)(8)(9)}=\frac{1}{2*7}-\frac{1}{8}+\frac{1}{2*9}............\)

\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)
\(2*29(\frac{1}{2*5}-\frac{1}{6}+\frac{1}{2*7}+\frac{1}{2*6}-\frac{1}{7}+\frac{1}{2*8}+\frac{1}{2*7}-\frac{1}{8}+\frac{1}{2*9}............\)

Terms will get cancelled out => \(\frac{1}{2*7}-\frac{1}{7}+\frac{1}{2*7}=0\)
\(2*29(\frac{1}{2*5}-\frac{1}{2*6}-\frac{1}{2*29}+\frac{1}{2*30})\)
\(29(\frac{1}{5}-\frac{1}{6}-\frac{1}{29}+\frac{1}{30})\)
\(29(\frac{1}{5}-\frac{1}{6}+\frac{1}{30})-\frac{29}{29}\)
\(29(\frac{6-5+1}{30})-\frac{29}{29}=\frac{2*29}{30}-1=\frac{28}{30}=\frac{14}{15}\)
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Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 21 Dec 2023, 22:21
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Bunuel do you have similar questions for practice?
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Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 11 Jan 2025, 00:13
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liviofrol
Given that \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\), which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ ... +\frac{(2)(29)}{(28)(29)(30)}\)?

(A) \(\frac{14}{15}\)

(B) \(\frac{2}{35}\)

(C) \(\frac{117}{35}\)

(D) \(\frac{232}{35}\)

(E) \(\frac{232}{969}\)

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It is a Series question in which they have given you how to break down the term to make the question easier. Given:

\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ ... +\frac{(2)(29)}{(28)(29)(30)}\)

Take 29 common:

\(29 * [\frac{(2)}{(5)(6)(7)}+\frac{(2)}{(6)(7)(8)}+\frac{(2)}{(7)(8)(9)}+ ... +\frac{(2)}{(28)(29)(30)}]\)

Each of these terms can be broken down as per the given relation:

\(\frac{(2)}{(5)(6)(7)} = \frac{1}{(5)(6)} - \frac{1}{(6)(7)}\)

\(\frac{(2)}{(6)(7)(8)} = \frac{1}{(6)(7)} - \frac{1}{(7)(8)}\)

Note the pattern. When we add them all, 1/6*7 will get cancelled off. Similarly, all terms will get cancelled off except the first and last terms. We will get

\(29 * [\frac{1}{(5)(6)} - \frac{1}{(29)(30)}] = 29 * \frac{29 - 1}{29*30} = \frac{14}{15}\)

Answer (A)

Here is a discussion on how to solve Series Questions: https://youtu.be/KX8WNiyNUIo
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Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 22 Jul 2026, 04:14
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A telescoping sum is a sum in which most terms cancel each other, leaving only a few terms at the beginning and the end. It is called "telescoping" because the expression collapses like a handheld telescope.

---

Consider the simple example (10 - 9) + (9 - 8) + (8 - 7) + (7 - 6). Removing the brackets gives 10 - 9 + 9 - 8 + 8 - 7 + 7 - 6. The -9 cancels with +9, the -8 cancels with +8, and the -7 cancels with +7. Everything in the middle disappears, leaving only 10 - 6 = 4.

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The same idea applies to fractions. A very common identity is 2/[n(n+1)(n+2)] = 1/[n(n+1)] - 1/[(n+1)(n+2)]. If you substitute consecutive values of n, such as 5, 6, 7, ..., each term becomes "current fraction - next fraction." For example, you get (1/(5*6) - 1/(6*7)) + (1/(6*7) - 1/(7*8)) + (1/(7*8) - 1/(8*9)) + ... . Notice that every negative fraction is canceled by the identical positive fraction in the next term. After all the cancellations, only the first fraction and the last negative fraction remain. The final result is 1/(5*6) - 1/(29*30).

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A high scorer recognizes a telescoping sum by looking for the pattern A(n) - A(n+1). As soon as every term can be rewritten in this form, there is no need to add every term individually because all the middle terms will cancel automatically. The only terms that survive are the very first positive term and the very last negative term.

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Common telescoping patterns to memorize are: 1/n - 1/(n+1), 2/[n(n+1)(n+2)] = 1/[n(n+1)] - 1/[(n+1)(n+2)], and 1/(2^n) - 1/(2^(n+1)). In each case, the expression has the form "current term - next term," so nearly everything cancels.

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Exam takeaway: Whenever you see a long sum of consecutive terms, first ask yourself whether each term can be rewritten as "something - the next something." If the answer is yes, stop trying to add all the fractions. Think "telescoping = cancellation." This simple pattern recognition often reduces a long calculation to just the first term minus the last term.
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