Let n = 0, 1, 2, 3, 4, 5, 6, 7....
When n = 0 then 0^2 + 0 = 0. n = 1 we have 1^2 + 1 = 2. And when n = 2 we have 2^2 + 2 = 6. When n= 3 we have 3^2 + 3 = 12. When n = 4 we have 4^2 + 4 = 20. When n = 5 we have 5^2 + 5 = 30. When n = 6 = 6^2 + 6 = 42. And finally when n = 7 we have 7^2 + 7 = 56. So at n = 1, 2, ...7, ... Our values are = 2, 6, 12, 20, 30, 42, and 56. It is obvious that n is always an even number. Hence n^2 + n is always an even integer for all positive integers.
When n = -1 we have (-1)^2 - 1 = 0 when n = -2 we have (-2)^2 -2 = 2. When n = -3 we have (-3)^2 - 3 = 6. When n = -4 we have (-4)^2 - 4 = 16 - 4 =12. When n =-5 we have (-5)^2 -5 = 20. When n = -6 we have (-6)^2 - 6 = 30. When n = (-7)^2 - 7 = 42. Hence n^2 + n is always even for all integers
Answer: 196 m
Step-by-step explanation:
Answer:
71
Step-by-step explanation:
vertical sides are congruent
x+30=101
x=71
