Question

The sum of the squares of 3 consecutive positive integers is 116.

Answer 1

a+(a+1)+(a+2)=116

Answer 2

oops

Answer 3

a^2+(a+1)^2+(a+2)^2=116

Answer 4

hm it doesn't seem to work though. a is not an integer =\

Answer 5

These are the options that it gave me
3n2+ 5 = 116
3n2+ 3n + 3 = 116
3n2+ 6n + 5 = 116

Answer 6

3n2+ 6n + 5 = 116

Answer 7

this is weird though since if you solve for x, x won't be an integer...

Answer 8

(well "n" lol sorry i changed variables)

Answer 9

how did you get that 3n2+6n+5=116? that's what i don't know how to get by
myself

Answer 10

\(n^2+(n+1)^2+(n+2)^2\)
\(=n^2+(n^2+2n+1)+(n^2+4n+4)\)
\(=3n^2+6n+5\)

Answer 11

expand everything and combine like-terms

Answer 12

oh, haha I didn't see that before lol thanks ^-^

Answer 13

:)