An algebra problem by Hana Wehbi

Algebra Level 3

1 2 ! + 2 3 ! + 3 4 ! + ⋯ + 99 100 ! = 1 − 1 A ! \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{99}{100!} = 1 - \frac{1}{A!}

What is A A ?

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


The answer is 100.

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Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Marco Brezzi
Aug 18, 2017

Relevant wiki: Telescoping series

We have to find

S = 1 2 ! + 2 3 ! + … + 99 100 ! = ∑ n = 1 99 n ( n + 1 ) ! S=\dfrac{1}{2!}+\dfrac{2}{3!}+\ldots +\dfrac{99}{100!}=\displaystyle\sum_{n=1}^{99}\dfrac{n}{(n+1)!}

S = ∑ n = 1 99 n ( n + 1 ) ! = ∑ n = 1 99 n + 1 − 1 ( n + 1 ) ! = ∑ n = 1 99 1 n ! − 1 ( n + 1 ) ! = 1 1 ! − 1 ( 99 + 1 ) ! = 1 − 1 100 ! = 1 − 1 A ! \begin{aligned} S&=\sum_{n=1}^{99}\dfrac{n}{(n+1)!}\\ &=\sum_{n=1}^{99}\dfrac{n+1-1}{(n+1)!}\\ &=\sum_{n=1}^{99}\dfrac{1}{n!}-\dfrac{1}{(n+1)!}\\ &=\dfrac{1}{1!}-\dfrac{1}{(99+1)!}=1-\dfrac{1}{100!}=1-\dfrac{1}{A!} \end{aligned}

⟹ A = 100 \Longrightarrow A=\boxed{100}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice and correct. Thank you for sharing it.

Hana Wehbi - 3 years, 9 months ago

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