Answer:
proof below
Step-by-step explanation:
Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)
Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;
(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2
= 2(2a^2 + 2a + 2b^2 + 2b + 1)
Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.
9 squared is 81
81=9^2
Or, (3^2)^2. Double square!
Great Job! they are all correct. :)
Good luck in your next tests.
So in your given pattern, you need to find first the derivatives and observe the patter that occurs in the given functions. So with this kind of pattern, every fourth one is the same; that makes the 114th derivative is the same as the second derivative. It is known since 114/4 has a remainder of two
