Easier than expected

Algebra Level 2

What is the square root of

1 + 3 + 5 + 7 + 9 + 11 + 1789 ? 1 + 3 + 5 + 7 + 9 + 11 + \ldots 1789 ?


The answer is 895.

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.

2 solutions

Sum of the series of consecutive odd numbers starting from 1 is given by .. { L a s t n u m b e r + 1 2 } 2 \{ \dfrac {Last number+1} {2} \}^2
Therefore the root is
1789 + 1 2 = 895 \\~\\ \dfrac{1789+1}{2}=895
This is also the equal to the avrage * numbers in the arithmetic series.


Shishir G.
Sep 25, 2014

Sum of 1st n odd numbers is n^2 .. Given sequence contains 895 numbers.. Hence square root of sum is 895

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...