Frightening Factorial

Algebra Level 3

3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + 5 3 ! + 4 ! + 5 ! + ⋯ + 10005 10003 ! + 10004 ! + 10005 ! = 1 λ ! − 1 μ ! \small \frac{3}{1!+2!+3!}+ \frac{4}{2!+3!+4!} + \frac{5}{3!+4!+5!} +\cdots + \frac{10005}{10003!+10004!+10005!} = \frac{1}{\lambda!} - \frac{1}{\mu!}

If the equation above holds true for some positive integers λ \lambda and μ \mu , find λ + μ \lambda + \mu .

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


The answer is 10007.

Dhanvanth Balakrishnan by Dhanvanth Balakrishnan , 22, India

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2 solutions

Presenting @Dhanvanth Balakrishnan 's solution in another way.

S = ∑ k = 1 10003 k + 2 k ! + ( k + 1 ) ! + ( k + 2 ) ! = ∑ k = 1 10003 k + 2 k ! ( 1 + k + 1 ) + ( k + 2 ) ! = ∑ k = 1 10003 k + 2 k ! ( k + 2 ) + ( k + 2 ) ! = ∑ k = 1 10003 k + 2 ( k + 2 ) ( k ! + ( k + 1 ) ! ) = ∑ k = 1 10003 1 k ! + ( k + 1 ) ! = ∑ k = 1 10003 1 k ! ( 1 + k + 1 ) = ∑ k = 1 10003 1 k ! ( k + 2 ) = ∑ k = 1 10003 k + 1 k ! ( k + 1 ) ( k + 2 ) = ∑ k = 1 10003 k + 1 ( k + 2 ) ! = ∑ k = 1 10003 k + 2 − 1 ( k + 2 ) ! = ∑ k = 1 10003 ( 1 ( k + 1 ) ! − 1 ( k + 2 ) ! ) = 1 2 ! − 1 10005 ! \begin{aligned} S & = \sum_{k=1}^{10003} \frac {k+2}{k!+(k+1)!+(k+2)!} \\ & = \sum_{k=1}^{10003} \frac {k+2}{k!(1+k+1)+(k+2)!} \\ & = \sum_{k=1}^{10003} \frac {k+2}{k!(k+2)+(k+2)!} \\ & = \sum_{k=1}^{10003} \frac {k+2}{(k+2)(k!+(k+1)!)} \\ & = \sum_{k=1}^{10003} \frac 1{k!+(k+1)!} \\ & = \sum_{k=1}^{10003} \frac 1{k!(1+k+1)} \\ & = \sum_{k=1}^{10003} \frac 1{k!(k+2)} \\ & = \sum_{k=1}^{10003} \frac {k+1}{k!(k+1)(k+2)} \\ & = \sum_{k=1}^{10003} \frac {k+1}{(k+2)!} \\ & = \sum_{k=1}^{10003} \frac {k+2-1}{(k+2)!} \\ & = \sum_{k=1}^{10003} \left(\frac 1{(k+1)!} - \frac 1{(k+2)!} \right) \\ & = \frac 1{2!} - \frac 1{10005!} \end{aligned}

⟹ λ + μ = 2 + 10005 = 10007 \implies \lambda + \mu = 2+10005 = \boxed{10007}

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First,let us find the general term of the series.

So,the general term is

t n t_{n} = n + 2 n ! + ( n + 1 ) ! + ( n + 2 ) ! \frac{n+2}{n!+(n+1)!+(n+2)!}

Let us simplify the general term,

t n t_{n} = n + 2 n ! ( 1 + ( n + 1 ) + ( n + 2 ) ( n + 1 ) ) \frac{n+2}{n!(1+(n+1)+(n+2)(n+1))}

t n t_{n} = n + 2 n ! ( n + 2 + ( n + 2 ) ( n + 1 ) ) \frac{n+2}{n!(n+2+(n+2)(n+1))}

t n t_{n} = n + 2 n ! ( n + 2 ) ( n + 2 ) \frac{n+2}{n!(n+2)(n+2)}

t n t_{n} = 1 n ! ( n + 2 ) \frac{1}{n!(n+2)}

Multiplying the numerator and denominator with n+1,

t n t_{n} = n + 1 ( n + 2 ) ! \frac{n+1}{(n+2)!}

t n t_{n} = n + 2 − 1 ( n + 2 ) ! \frac{n+2-1}{(n+2)!}

t n t_{n} = 1 ( n + 1 ) ! \frac{1}{(n+1)!} - 1 ( n + 2 ) ! \frac{1}{(n+2)!}

Now , we can substitute the values of n in the general term.

= 1 2 ! \frac{1}{2!} - 1 3 ! \frac{1}{3!} + 1 3 ! \frac{1}{3!} - 1 4 ! \frac{1}{4!} . . . ... - 1 10005 ! \frac{1}{10005!}

= 1 2 ! \frac{1}{2!} - 1 10005 ! \frac{1}{10005!}

So,

λ λ = 2 2

μ μ = 10005 10005

Therefore,

λ λ + μ μ = 10007 \boxed{10007}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

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