Math Series #7

Number Theory Level 2

Calculate ( 2 2 + 4 2 + 6 2 + . . . + 10 0 2 ) − ( 1 2 + 3 2 + 5 2 + . . . + 9 9 2 ) (2^{2} + 4^{2} + 6^{2} + ... + 100^{2}) - (1^{2} + 3^{2} + 5^{2} + ... + 99^{2}) .


The answer is 5050.

Avner Lim by Avner Lim , 38, Indonesia

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

Pop Wong
Mar 10, 2021

( 2 2 + 4 2 + . . . + 10 0 2 ) − ( 1 2 + 3 2 + . . . . + 9 9 2 ) (2^2+4^2+...+100^2) - (1^2 +3^2 +....+ 99^2)

= ∑ 1 50 ( 2 n ) 2 − ( 2 n − 1 ) 2 = ∑ 1 50 4 n − 1 = 3 + 7 + . . . + 199 = 50 ( 3 + 199 ) 2 = 50 × 101 = 5050 = \sum_{1}^{50} (2n)^2 - (2n-1)^2 \\ = \sum_{1}^{50} 4n - 1 \\ = 3 + 7 + ... + 199 \\ = 50\cfrac{(3+199)}{2} \\ = 50\times101 \\ = \boxed{5050}

0 Helpful 1 Interesting 2 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...