Simple series Relation

Algebra Level 3

If x = 1 1 2 + 1 3 2 + 1 5 2 . . . . . x\quad = \frac { 1 }{ 1^ 2 } +\frac { 1 }{ 3^ 2 } +\frac { 1 }{ 5^ 2 } ..... ,,,,, y = 1 1 2 + 3 2 2 + 1 3 2 + 3 4 2 + . . . . y\quad =\quad \frac { 1 }{ 1^ 2 } +\frac { 3 }{ 2^ 2 } +\frac { 1 }{ 3^ 2 } +\frac { 3 }{ 4^ 2 } +.... , z = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + . . . . z\quad =\quad \frac { 1 }{ 1^ 2 } -\frac { 1 }{ 2^ 2 } +\frac { 1 }{ 3^ 2 } -\frac { 1 }{ 4^ 2 } +.... , then which of the following is true ?

y 6 , x 3 , z \dfrac{y}{6}, \dfrac{x}{3}, z are in A.P. 6 y , 3 x , 2 z 6y,3x,2z are in A.P. y 6 , x 3 , z 2 \dfrac{y}{6}, \dfrac{x}{3}, \dfrac{z}{2} are in A.P. x , y , z x, y, z are in A.P.

Pankaj Joshi by Pankaj Joshi , 23, India

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1 solution

Ronak Agarwal
Jul 24, 2014

T a k e a = ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 a n d b = ∑ n = 1 ∞ 1 ( 2 n ) 2 . S o w e h a v e x = a y = a + 3 b z = a − b H e n c e w e h a v e y 6 + z 2 = a + 3 b 6 + a − b 2 = 2 a 3 2 ( x 3 ) . ⇒ y / 6 , x / 3 , z / 2 a r e i n A P Take\quad a=\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2n-1) }^{ 2 } } } \quad and\quad b=\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2n })^{ 2 } } } .\\ So\quad we\quad have\quad x=a\quad \quad y=a+3b\quad \quad z=a-b\\ Hence\quad we\quad have\quad \frac { y }{ 6 } +\frac { z }{ 2 } =\frac { a+3b }{ 6 } +\frac { a-b }{ 2 } =\frac { 2a }{ 3 } 2(\frac { x }{ 3 } ).\\ \Rightarrow y/6,x/3,z/2\quad are\quad in\quad AP

5 Helpful 0 Interesting 0 Brilliant 0 Confused

You forgot the last equal sign. That caused me confusion at first. Very nice solution though! I don't think it can get any easier.

James Wilson - 3 years, 6 months ago

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