Very easy sumation (calculator allowed)

Algebra Level 4

If f ( n ) = 1 ( n + ( n + 1 ) ) 2 − n ( n + 1 ) , f(n) = \frac{1}{(n + (n + 1))^2 - n(n + 1)},

then find 1 f ( 1 ) + 1 f ( 2 ) + 1 f ( 3 ) + . . . + 1 f ( 99 ) . \frac { 1 }{ f(1) } +\frac { 1 }{ f(2) } +\frac { 1 }{ f(3) } +...+\frac { 1 }{ f(99) }.


The answer is 999999.

Porames Vattanaprasan by Porames Vattanaprasan , 21, Thailand

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2 solutions

Kerry Soderdahl
Sep 1, 2015

Let g ( x ) = 1 f ( x ) g(x)=\frac{1}{f(x)}

We are asked to find:

∑ n = 1 99 1 f ( n ) = ∑ n = 1 99 g ( n ) \sum_{n=1}^{99}\frac{1}{f(n)}=\sum_{n=1}^{99}g(n)

g ( x ) = ( x + ( x + 1 ) ) 2 − x ( x + 1 ) = ( 2 x + 1 ) 2 − x 2 − 1 g(x)= (x+(x+1))^2-x(x+1)=(2x+1)^2-x^2-1

= 4 x 2 + 4 x + 1 − x 2 − 1 = 3 x 2 + 3 x + 1 =4x^2+4x+1-x^2-1=3x^2+3x+1

= [ x 3 + 3 x 2 + 3 x + 1 ] − x 3 = ( x + 1 ) 3 − x 3 =[x^3+3x^2+3x+1]-x^3=(x+1)^3-x^3

I express g ( x ) g(x) as ( x + 1 ) 3 − x 3 (x+1)^3-x^3 to create what is called a telescoping sum , where successive terms will "cancel out" each other.

∑ n = 1 99 g ( n ) = ∑ n = 1 99 ( − n 3 + ( n + 1 ) 3 ) \sum_{n=1}^{99}g(n)=\sum_{n=1}^{99}(-n^3+(n+1)^3)

= − 1 3 + 2 3 − 2 3 + 3 3 − 3 3 + 4 3 − 4 3 ± … + 9 9 3 − 9 9 3 + 10 0 3 =-1^3+2^3-2^3+3^3-3^3+4^3-4^3\pm\ldots+99^3-99^3+100^3

= − 1 + 10 0 3 = 1000000 − 1 = 999999 =-1+100^3=1000000-1=\boxed{999999}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

F r o m f ( 1 ) = 1 ( 1 + 2 ) 2 − 2 L e t 1 = a a n d 2 = b f ( 1 ) = 1 ( a + b ) 2 − a b = 1 a 2 + a b + b 2 C o n j u g a t e ; 1 a 2 + a b + b 2 × a − b a − b = a − b a 3 − b 3 1 f ( 1 ) = a 3 − b 3 a − b = 1 3 − 2 3 − 1 = − 1 3 + 2 3 S o 1 f ( 1 ) + 1 f ( 2 ) + 1 f ( 3 ) + . . . + 1 f ( 99 ) = − 1 3 + 2 3 − 2 3 + 3 3 − 3 3 + 4 3 − . . . − 99 3 + 100 3 = − 1 3 + 100 3 = 999 , 999 From\quad f(1)\quad =\quad \frac { 1 }{ { (1+2) }^{ 2 }-2 } \\ Let\quad 1=a\quad and\quad 2=b\\ f(1)\quad =\quad \frac { 1 }{ { (a+b) }^{ 2 }-ab } \quad =\quad \frac { 1 }{ { a }^{ 2 }+ab{ +b }^{ 2 } } \\ Conjugate;\quad \frac { 1 }{ { a }^{ 2 }+ab{ +b }^{ 2 } } \times \frac { a-b }{ a-b } \quad =\quad \frac { a-b }{ { a }^{ 3 }-{ b }^{ 3 } } \quad \\ \frac { 1 }{ f(1) } \quad =\quad \frac { { a }^{ 3 }-{ b }^{ 3 } }{ { a }-{ b } } \quad =\quad \frac { 1^{ 3 }-{ 2 }^{ 3 } }{ -1 } \quad =\quad -{ 1 }^{ 3 }+{ 2 }^{ 3 }\\ So\quad \frac { 1 }{ f(1) } +\frac { 1 }{ f(2) } +\frac { 1 }{ f(3) } +...+\frac { 1 }{ f(99) } \quad =\quad { -1 }^{ 3 }+{ 2 }^{ 3 }-{ 2 }^{ 3 }+3^{ 3 }{ -3 }^{ 3 }+4^{ 3 }-...-{ 99 }^{ 3 }+{ 100 }^{ 3 }\quad =\quad { -1 }^{ 3 }+{ 100 }^{ 3 }\quad =\quad 999,999\\

1 Helpful 0 Interesting 0 Brilliant 0 Confused

In the question shouldn't the squared be outside of the bracket because I thought the first bit was n+ (n+1)^2 and not (n+(n+1))^2 and from the solution, I think you imply the latter.

Raghav Arora - 5 years, 9 months ago

Question is wrong

Riddhesh Deshmukh - 5 years, 8 months ago

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