Do Not Expand

Algebra Level 3

True or False?

For all positive integers k k , there are no {\color{#D61F06} \text{no}} real solutions to

a 2 + ( a + 1 ) 2 + ( a + 2 ) 2 + ⋯ + ( a + k ) 2 = 0 a^2 + (a+1)^2 + (a+2)^2 + \cdots + (a+k)^2 = 0

False True Depends on k k

Chung Kevin by Chung Kevin , 46, Australia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ayush G Rai
Oct 30, 2016

The sum of the squares should always be greater than or equal to 0. 0. Since it is equal to 0 0 ,each term also must be 0. 0. So, a 2 = 0 ⇒ a = 0. a^2=0\Rightarrow a=0.
But ( a + k ) 2 = 0 ⇒ ( k ) 2 = 0 ⇒ k = 0 {(a+k})^2=0\Rightarrow {(k)}^2=0\Rightarrow k=0 ,which is not possible since k k should be a positive integer.Hence there are no real solutions.

5 Helpful 1 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...