What is the value of \(27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2?\)

This topic has expert replies
Source: — Problem Solving |

Master | Next Rank: 500 Posts
Posts: 416
Joined: Thu Oct 15, 2009 11:52 am
Thanked: 27 times
Rewrite with 30 as the base:

(30-3)^2+(30-2)^2+(30-1)^2+30^2+ (30+1)^2+(30+2)^2+(30+3)^2.

Notice the second terms in each. Those will be squared, so will become 9,4,1,1,4,9. These total 28...😉.

Notice these same second terms will be applied twice against 30 if they were multiplied out. Since there are the same terms both positive and negative, they will cancel out.

Now we are just left with seven sets of 30^2.

30^2=900. 7x900=6300.

6300+28=6328,D