Telescoping the Reciprocals!

Algebra Level 4

3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + 5 3 ! + 4 ! + 5 ! + ⋯ + 100 98 ! + 99 ! + 100 ! \begin{aligned} \frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \frac{5}{3!+4!+5!} + \cdots+ \frac{100}{98!+99!+100!} \end{aligned}

Find the value of the expression above.

The answer is a form of 1 a ! − 1 b ! \frac{1}{a!} - \frac{1}{b!} . Submit your answer as a × b a \times b .

Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

This problem is one of my set, Let's Practice .


The answer is 200.

Fidel Simanjuntak by Fidel Simanjuntak , 19, Indonesia

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3 solutions

Relevant wiki: Telescoping Series - Sum

Now, note that 3 1 ! + 2 ! + 3 ! = 3 1 ! ( 1 + 2 + 2 × 3 ) \frac{3}{1!+2!+3!} = \frac{3}{1!(1+2+2 \times 3)}

3 1 ! × 9 = 1 3 = 2 3 × 2 × 1 \frac{3}{1! \times 9} = \frac{1}{3} = \frac{2}{3 \times 2 \times 1}

= 2 3 ! = 1 2 ! − 1 3 ! = \frac{2}{3!} = \frac{1}{2!} - \frac{1}{3!} .

We have 3 1 ! + 2 ! + 3 ! = 1 2 ! − 1 3 ! \frac{3}{1!+2!+3!} = \frac{1}{2!} - \frac{1}{3!}

Simplify each terms, we have

3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + 5 3 ! + 4 ! + 5 ! + . . . + 100 98 ! + 99 ! + 100 ! \frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \frac{5}{3!+4!+5!} + ... + \frac{100}{98!+99!+100!}

= 1 2 ! − 1 3 ! + 1 3 ! − 1 4 ! + 1 4 ! − 1 5 ! + . . . . . . . . − 1 100 ! = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \frac{1}{4!} - \frac{1}{5!} + ........ - \frac{1}{100!}

After all, we have

1 2 ! − 1 100 ! = 1 a ! − 1 b ! \frac{1}{2!} - \frac{1}{100!} = \frac{1}{a!} - \frac{1}{b!}

a = 2 a=2 and b = 100 b=100 .

Hence, a × b = 200 a \times b = \boxed{200} .

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Prakhar Bindal
Jan 3, 2017

The general term of the series is

1/r!(r+2)

Multiplying and dividing by r+1

r+1/(r+2)! = 1/(r+1)!-1/(r+2)!

which is clearly a telescoping series and telescopes to 1/2!-1/100!

Hence ab = 200

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Hi sir. I want to ask about something. How to put this problem into my set? I forget how to do it. Can you help me? - @Prakhar Bindal

Fidel Simanjuntak - 4 years, 5 months ago

Log in to reply

and you can call me prakhar!

Prakhar Bindal - 4 years, 5 months ago

At the bottom right of the problem Three dots will be coming . you click on that you will get the option save click on that @Fidel Simanjuntak

Prakhar Bindal - 4 years, 5 months ago

Ok, thx Prakhar.

Fidel Simanjuntak - 4 years, 5 months ago

a n = n ( n − 2 ) ! + ( n − 1 ) ! + ( n ) ! = n ( n − 2 ) ! ∗ { 1 + ( n − 1 ) + ( n − 1 ) ∗ n } = 1 ( n − 2 ) ! ∗ n = n − 1 ( n − 2 ) ! ∗ ( n − 1 ) ∗ n = n − 1 n ! = 1 ( n − 1 ) ! − 1 n ! ∴ ∑ n = 3 100 a n = ∑ n = 3 100 { 1 ( n − 1 ) ! − 1 n ! } ∴ ∑ n = 3 3 1 ( n − 1 ) ! + ∑ n = 4 100 1 ( n − 1 ) ! − ∑ n = 3 100 1 n ! ∴ ∑ n = 3 3 1 ( n − 1 ) ! + { ∑ n = 3 99 1 n ! − ∑ n = 3 99 1 n ! } − ∑ n = 100 100 1 n ! ∴ ∑ n = 3 100 a n = ∑ n = 3 3 1 ( n − 1 ) ! + 0 − ∑ n = 100 100 1 n ! = 1 2 ! − 1 100 ! = 1 a ! − 1 b ! ∴ a ∗ b = 2 ∗ 100 = 200. a_n=\dfrac n {(n-2)!+(n-1)!+(n)!}=\dfrac n {(n-2)!*\bigg \{1+(n-1)+(n-1)*n \bigg \} }\\ =\dfrac 1 {(n-2)!*n}=\dfrac{n-1}{(n-2)!*(n-1)*n}\\ =\dfrac {n-1}{n!}=\dfrac 1 {(n-1)!}- \dfrac 1 {n!}\\ \therefore~\displaystyle \sum_{n=3}^{100} a_n= \sum_{n=3}^{100} \Bigg \{\dfrac 1 {(n-1)!}- \dfrac 1 {n!} \Bigg \}\\ \therefore~\displaystyle \sum_{n=3}^3 \dfrac 1 {(n-1)!} +\sum_{n=4}^{100} \dfrac 1 {(n-1)!} - \sum_{n=3}^{100} \dfrac 1 {n!}\\ \therefore~\displaystyle \sum_{n=3}^3 \dfrac 1 {(n-1)!} + { \Large \Bigg \{ \color{#3D99F6}{ \sum_{n=3}^{99} \dfrac 1 {n!} - \sum_{n=3}^{99} \dfrac 1 {n!}} \Bigg \} } - \sum_{n=100}^{100} \dfrac 1 {n!} \\ \therefore~\displaystyle \sum_{n=3}^{100} a_n=\sum_{n=3}^3 \dfrac 1 {(n-1)!} + 0 - \sum_{n=100}^{100} \dfrac 1 {n!} \\ =\dfrac 1 {2!} - \dfrac 1 {100!}=\dfrac 1 {a!} - \dfrac 1 {b!}\\ \therefore~a*b=2*100=\Large \color{#D61F06}{200}.

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