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 .

100 99 3 2

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Hana Wehbi by Hana Wehbi , 51, USA

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2 solutions

Similar solution to Aruna Addaguduru 's

S = 1 2 ! + 2 3 ! + 3 4 ! + . . . + 99 100 ! = ∑ k = 2 100 k − 1 k ! = ∑ k = 2 100 ( k k ! − 1 k ! ) = ∑ k = 2 100 ( 1 ( k − 1 ) ! − 1 k ! ) = 1 1 ! − 1 100 ! = 1 − 1 100 ! \begin{aligned} S & = \frac 1{2!} + \frac 2{3!} + \frac 3{4!} + ... + \frac {99}{100!} \\ & = \sum_{k=2}^{100} \frac {k-1}{k!} \\ & = \sum_{k=2}^{100} \left( \frac k{k!} - \frac 1{k!} \right) \\ & = \sum_{k=2}^{100} \left( \frac 1{(k-1)!} - \frac 1{k!} \right) \\ & = \frac 1{1!} - \frac 1{100!} \\ & = 1 - \frac 1{100!} \end{aligned}

⟹ A = 100 \implies A=\boxed{100}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. Thank you.

Hana Wehbi - 4 years, 3 months ago
Sudhamsh Suraj
Mar 8, 2017

General term of the given expression is n ( n + 1 ) ! \frac{n}{(n+1)!}

So writing numerator 'n' as (n+1)-(1) and simplyfying

We get general term as 1 n ! \frac{1}{n!} - 1 ( n + 1 ) ! \frac{1}{(n+1)!}

So adding all terms i.e. from n=1 to n=99

We get all terms cancelled except 1st and last term

Which is equal to 1 - 1 100 ! \frac{1}{100!}

Therefore, A=100.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. Thank you

Hana Wehbi - 4 years, 3 months ago

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