An algebra problem by Wildan Bagus Wicaksono

Algebra Level 3

f ( n ) = 4 n + 4 n 2 − 1 2 n + 1 + 2 n − 1 f(n)=\frac { 4n+\sqrt { 4{ n }^{ 2 }-1 } }{ \sqrt { 2n+1 } +\sqrt { 2n-1 } }

Function f : N → R f : \mathbb {N \to R} is defined as above. Determine f ( 13 ) + f ( 14 ) + f ( 15 ) + . . . + f ( 112 ) f(13)+f(14)+f(15)+...+f(112) .


The answer is 1625.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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1 solution

f ( n ) = 4 n + 4 n 2 − 1 2 n + 1 + 2 n − 1 f ( n ) = 4 n + 4 n 2 − 1 2 n + 1 + 2 n − 1 ⋅ 2 n + 1 − 2 n − 1 2 n + 1 − 2 n − 1 f ( n ) = ( 4 n + 4 n 2 − 1 ) ( 2 n + 1 − 2 n − 1 ) ( 2 n + 1 ) − ( 2 n − 1 ) f ( n ) = 1 2 { [ ( 2 n + 1 ) + ( 2 n − 1 ) + 2 n + 1 ⋅ 2 n − 1 ] ( 2 n + 1 − 2 n − 1 ) } f ( n ) = 1 2 { ( 2 n + 1 ) 2 n + 1 − ( 2 n − 1 ) 2 n − 1 } f(n)=\frac { 4n+\sqrt { 4{ n }^{ 2 }-1 } }{ \sqrt { 2n+1 } +\sqrt { 2n-1 } } \\ f(n)=\frac { 4n+\sqrt { 4{ n }^{ 2 }-1 } }{ \sqrt { 2n+1 } +\sqrt { 2n-1 } } \cdot \frac { \sqrt { 2n+1 } -\sqrt { 2n-1 } }{ \sqrt { 2n+1 } -\sqrt { 2n-1 } } \\ f(n)=\frac { \left( 4n+\sqrt { 4{ n }^{ 2 }-1 } \right) \left( \sqrt { 2n+1 } -\sqrt { 2n-1 } \right) }{ (2n+1)-(2n-1) } \\ f(n)=\frac { 1 }{ 2 } \left\{ \left[ (2n+1)+(2n-1)+\sqrt { 2n+1 } \cdot \sqrt { 2n-1 } \right] \left( \sqrt { 2n+1 } -\sqrt { 2n-1 } \right) \right\} \\ f(n)=\frac { 1 }{ 2 } \left\{ (2n+1)\sqrt { 2n+1 } -(2n-1)\sqrt { 2n-1 } \right\}

f ( 13 ) = 1 2 { 27 27 − 25 25 } f ( 14 ) = 1 2 { 29 29 − 27 27 } f ( 15 ) = 1 2 { 31 31 − 29 29 } … f ( 112 ) = 1 2 { 225 225 − 223 223 } f(13)=\frac { 1 }{ 2 } \left\{ 27\sqrt { 27 } -25\sqrt { 25 } \right\} \\ f(14)=\frac { 1 }{ 2 } \left\{ 29\sqrt { 29 } -27\sqrt { 27 } \right\} \\ f(15)=\frac { 1 }{ 2 } \left\{ 31\sqrt { 31 } -29\sqrt { 29 } \right\} \\ \dots \\ f(112)=\frac { 1 }{ 2 } \left\{ 225\sqrt { 225 } -223\sqrt { 223 } \right\}

f ( 13 ) + f ( 14 ) + f ( 15 ) + . . . + f ( 112 ) = 1 2 { 225 225 − 25 25 } f ( 13 ) + f ( 14 ) + f ( 15 ) + . . . + f ( 112 ) = 1 2 { 225 ⋅ 15 − 25 ⋅ 5 } f ( 13 ) + f ( 14 ) + f ( 15 ) + . . . + f ( 112 ) = 1 2 ( 3375 − 125 ) f ( 13 ) + f ( 14 ) + f ( 15 ) + . . . + f ( 112 ) = 1625 f(13)+f(14)+f(15)+...+f(112)=\frac { 1 }{ 2 } \left\{ 225\sqrt { 225 } -25\sqrt { 25 } \right\} \\ f(13)+f(14)+f(15)+...+f(112)=\frac { 1 }{ 2 } \left\{ 225\cdot 15-25\cdot 5 \right\} \\ f(13)+f(14)+f(15)+...+f(112)=\frac { 1 }{ 2 } (3375-125)\\ f(13)+f(14)+f(15)+...+f(112)=1625

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