This is not what you think!

Algebra Level 4

3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + ⋯ + 2016 2014 ! + 2015 ! + 2016 ! \large\dfrac{3}{1!+2!+3!} + \dfrac{4}{2!+3!+4!} +\cdots+ \dfrac{2016}{2014!+2015!+2016!}

If the above expression can be represented as 1 k ! − 1 ( k + a ) ! , \dfrac{1}{k!} - \dfrac{1}{(k+a)!} \; , then find a k \dfrac{a}{k} .

Notation :

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


The answer is 1007.

Arsan Safeen by Arsan Safeen , 23, Iraq

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1 solution

Arsan Safeen
Apr 3, 2016

3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + . . . . . . . . . . + 2016 2014 ! + 2015 ! + 2015 ! \frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + .......... + \frac{2016}{2014!+2015!+2015!}

k ( k − 2 ) ! + ( k − 1 ) ! + k ! \frac{k}{(k-2)!+(k-1)!+k!}

= k ( k − 2 ) ! ( 1 + ( k − 1 ) + k ( k − 1 ) ) =\frac{k}{(k-2)!(1+(k-1)+k(k-1))}

= k ( k − 2 ) ! ( k 2 ) =\frac{k}{(k-2)!(k^{2})}

= 1 ( k − 2 ) ! ( k ) =\frac{1}{(k-2)!(k)}

= k − 1 k ! =\frac{k-1}{k!}

= 1 ( k − 2 ) ! ( k ) =\frac{1}{(k-2)!(k)}

= 1 ( k − 1 ) ! − 1 k ! =\frac{1}{(k-1)!} - \frac{1}{k!} ,then

1 2 ! − 1 3 ! + 1 3 ! − 1 4 ! + 1 4 ! − 1 5 ! + . . . . . . . . . . + 1 2015 ! − 1 2016 ! \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \frac{1}{4!} - \frac{1}{5!} + .......... + \frac{1}{2015!} - \frac{1}{2016!}

= 1 2 ! − 1 2016 ! =\frac{1}{2!} - \frac{1}{2016!}

= 1 2 ! − 1 ( 2014 + 2 ) ! =\frac{1}{2!} - \frac{1}{(2014+2)!}

then, a = 2014 a = 2014 and k = 2 k = 2

Therefore, a k = 2014 2 = 1007 \frac{a}{k} = \frac{2014}{2} = 1007

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