Another Factorial Summation! (CMI 2010)

Algebra Level 3

2 0 ! + 1 ! + 2 ! + 3 1 ! + 2 ! + 3 ! + ⋯ + 100 98 ! + 99 ! + 100 ! \dfrac2{0! +1!+2!} + \dfrac3{1!+2!+3!} + \cdots+ \dfrac{100}{98!+99!+100!}

If the value of the expression above is equal to x x , evaluate x + 1 100 ! x + \dfrac1{100!} .


The answer is 1.

Aditya Dhawan by Aditya Dhawan , 20, India

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1 solution

Let’s rewrite it in summation notation as - \text{Let's rewrite it in summation notation as -}

S = ∑ k = 0 98 k + 2 k ! + ( k + 1 ) ! + ( k + 2 ) ! \large \mathfrak{S}=\sum_{k=0}^{98} \frac{k+2}{k!+(k+1)!+(k+2)!}

S = ∑ k = 0 98 k + 2 k ! + ( k + 1 ) k ! + ( k + 2 ) ( k + 1 ) k ! \mathfrak{S}=\sum_{k=0}^{98} \frac{k+2}{k!+(k+1)k!+(k+2)(k+1)k!}

S = ∑ k = 0 98 k + 2 k ! ( k + 2 + ( k + 1 ) ( k + 2 ) ) \mathfrak{S}=\sum_{k=0}^{98} \frac{k+2}{k!(k+2 + (k+1)(k+2))}

S = ∑ k = 0 98 k + 2 k ! ( k + 2 ) 2 ) \mathfrak{S}=\sum_{k=0}^{98} \frac{k+2}{k!(k+2)^2)}

S = ∑ k = 0 98 1 k ! ( k + 2 ) ) \mathfrak{S}=\sum_{k=0}^{98} \frac{1}{k!(k+2))}

S = ∑ k = 0 98 k + 1 k ! ( k + 2 ) ( k + 1 ) ) \mathfrak{S}=\sum_{k=0}^{98} \frac{k+1}{k!(k+2)(k+1))}

S = ∑ k = 0 98 k + 2 − 1 ( k + 2 ) ! \mathfrak{S}=\sum_{k=0}^{98} \frac{k+2-1}{(k+2)!}

S = ∑ k = 0 98 k + 2 ( k + 2 ) ! − 1 ( k + 2 ) ! \mathfrak{S}=\sum_{k=0}^{98} \frac{k+2}{(k+2)!}-\frac{1}{(k+2)!}

S = ∑ k = 0 98 1 ( k + 1 ) ! − 1 ( k + 2 ) ! \mathfrak{S}=\sum_{k=0}^{98} \frac{1}{(k+1)!}-\frac{1}{(k+2)!}

Now the above is a telescoping series and looks as follows : \text{Now the above is a telescoping series and looks as follows :}

S = 1 1 ! − 1 2 ! + 1 2 ! − 1 3 ! + 1 3 ! − ⋯ + 1 99 ! − 1 100 ! \large \color{#624F41}{\mathfrak{S} = \frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\cdots+\frac{1}{99!}-\frac{1}{100!}}

S = 1 − 1 100 ! \large \mathfrak{S} = 1-\frac{1}{100!}

So , S + 1 100 ! = 1 \text{So} , \mathfrak{S} + \frac{1}{100!} =\boxed{1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Same approach :) (+1)

In the last line it should be 1 100 ! \frac{1}{100!} not 1 100 \frac{1}{100}

Aditya Dhawan - 5 years, 2 months ago

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Thanks! I hve edited that

Aditya Narayan Sharma - 5 years, 2 months ago

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