Step-by-step explanation:
many solutions
1st number = n
2nd number = n+1
3rd number = n+2
sum of the squares of 3 consecutive numbers is 116
n² + (n+1)² + (n+2)² = 116
n² + (n+1)(n+1) + (n+2)(n+2) = 116
n² + [n(n+1)+1(n+1)] + [n(n+2)+2(n+2)] = 116
n² + n² + n + n + 1 + n² + 2n + 2n + 4 = 116
n² + n² + n² + n + n + 2n + 2n + 1 + 4 = 116
3n² + 6n + 5 = 116 Last option.
Answer:
Odd integers
Step-by-step explanation:
The range is odd integers, if you add 1 to any even integer it becomes an odd integer. 2+1=3; 10+1=11
Its 234!!!!!!!!!!!!!! Just checked my work
