An algebra problem by Wildan Bagus Wicaksono

Algebra Level 3

Let

x = 3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + 5 3 ! + 4 ! + 5 ! + . . . + 2017 2015 ! + 2016 ! + 2017 ! x=\frac { 3 }{ 1!+2!+3! } +\frac { 4 }{ 2!+3!+4! } +\frac { 5 }{ 3!+4!+5! } +...+\frac { 2017 }{ 2015!+2016!+2017! }

Determine the value of x + 1 2017 ! x+\dfrac 1{2017!} and write it in 3 decimal places.


The answer is 0.5.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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1 solution

= 3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + 5 3 ! + 4 ! + 5 ! + . . . + 2017 2015 ! + 2016 ! + 2017 ! = ∑ k = 1 98 k + 2 k ! + ( k + 1 ) ! + ( k + 2 ) ! = ∑ k = 1 98 1 k ! { k + 2 1 + ( k + 1 ) + ( k + 1 ) ( k + 2 ) } = ∑ k = 1 98 1 k ! { k + 2 k + 2 + k 2 + 3 k + 2 } = ∑ k = 1 98 1 k ! { k + 2 k 1 + 4 k + 4 } = ∑ k = 1 98 1 k ! { k + 2 ( k + 2 ) 2 } = ∑ k = 1 98 1 k ! { 1 k + 2 } = ∑ k = 1 98 1 k ! ( k + 2 ) = ∑ k = 1 98 k + 1 k ! ( k + 1 ) ( k + 2 ) = ∑ k = 1 98 k + 2 − 1 ( k + 2 ) ! = ∑ k = 1 98 k + 2 ( k + 2 ) ! − 1 ( k + 2 ) ! = ∑ k = 1 98 1 ( k + 1 ) ! − 1 ( k + 2 ) ! \large =\frac { 3 }{ 1!+2!+3! } +\frac { 4 }{ 2!+3!+4! } +\frac { 5 }{ 3!+4!+5! } +...+\frac { 2017 }{ 2015!+2016!+2017! } \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { k+2 }{ k!+(k+1)!+(k+2)! } } \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { 1 }{ k! } } \left\{ \frac { k+2 }{ 1+(k+1)+(k+1)(k+2) } \right\} \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { 1 }{ k! } } \left\{ \frac { k+2 }{ k+2+{ k }^{ 2 }+3k+2 } \right\} \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { 1 }{ k! } } \left\{ \frac { k+2 }{ { k }^{ 1 }+4k+4 } \right\} \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { 1 }{ k! } } \left\{ \frac { k+2 }{ { (k+2) }^{ 2 } } \right\} \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { 1 }{ k! } } \left\{ \frac { 1 }{ k+2 } \right\} \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { 1 }{ k!(k+2) } } \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { k+1 }{ k!(k+1)(k+2) } } \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { k+2-1 }{ (k+2)! } } \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { k+2 }{ (k+2)! } } -\frac { 1 }{ (k+2)! } \\ =\large \sum _{ k=1 }^{ 98 }{ \frac { 1 }{ (k+1)! } -\frac { 1 }{ (k+2)! } }

x = 3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + 5 3 ! + 4 ! + 5 ! + . . . + 2017 2015 ! + 2016 ! + 2017 ! x = 1 2 ! − 1 3 ! + 1 3 ! − 1 4 ! + . . . + 1 2016 ! − 1 2017 ! x = 1 2 ! − 1 2017 ! x + 1 2017 ! = 1 2 ! − 1 2017 ! + 1 2017 ! x + 1 2017 ! = 1 2 x + 1 2017 ! = 0.5 \large x=\frac { 3 }{ 1!+2!+3! } +\frac { 4 }{ 2!+3!+4! } +\frac { 5 }{ 3!+4!+5! } +...+\frac { 2017 }{ 2015!+2016!+2017! } \\ \large x=\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\frac { 1 }{ 3! } -\frac { 1 }{ 4! } +...+\frac { 1 }{ 2016! } -\frac { 1 }{ 2017! } \\ \large x=\frac { 1 }{ 2! } -\frac { 1 }{ 2017! } \\ \large x+\frac { 1 }{ 2017! } =\frac { 1 }{ 2! } -\frac { 1 }{ 2017! } +\frac { 1 }{ 2017! } \\ \large x+\frac { 1 }{ 2017! } =\frac { 1 }{ 2 } \\ \large x+\frac { 1 }{ 2017! } =0.5

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