I need help...if you can please help me with a simple but difficult problem, it would be greatly appreciated.?

Question

Provide a counter exampleforc each statement.

1.(n+2)3 =n3 +2 3 (I don't know how to make the numbers smaller, sorry)

Answer:

2. x2 > x for all values of x.

Answer:

Tell whether each statement is tru or false. if it is false, give a counterexample.

3.For every positive integer n,n2 +n +11 is a prime number.

Answer:


4.The result of multiplying the sum of the first n postive even integers by 4 and then adding 1 is always a perfect square.

Answer:

Answer 1

1. FALSE. Take n = 2. Then (2+2)³ = 64 while 2³ + 2³ =16.
This kind of error is often called the "Freshman mistake".
2. It's FALSE for x = 0. 0² = 0. It's also false for
0 < x <= 1.
3. FALSE. If n = 11, n² + n +11 is divisible by 11.
It also fails for n = 10: 10²+ 10 + 11 = 121 = 11².
4. TRUE. 2 + 4 + ... + 2n = 2(1+2 +...+ n)
= 2n(n+1)/2 = n(n+1).
If we multiply this by 4 and add 1 we get
4n²+4n+1 = (2n+1)².
Hope that helps!

Answer 2

Freshman's dream!

1.(n+2)^3 =n^3 +2^ 3 (I don't know how to make the numbers smaller, sorry)

Answer: How about n = 1?

(1+2)^3 = 27 <> 1^3 + 2^3 = 1+8, ie,
27 <> 9.

2. x2 > x for all values of x.

Answer: Well, 1^2 = 1, NOT strictly greater than 1.

Tell whether each statement is tru or false. if it is false, give a counterexample.

3.For every positive integer n,n^2 +n +11 is a prime number.

Answer: I think it's true, but then, I don't have the time to be sure.


4.The result of multiplying the sum of the first n postive even integers by 4 and then adding 1 is always a perfect square.

Answer: (1+2+3+ ...n)*4 + 1

S = n(n+1)/2.

Take n = 3.

S = 3*4/2 = 6. Add 4, = 10, NOT a perfect square!

Work on formula S = n(n+1)/2 * 4 + 1
= 2n(n+1) + 1
= 2n^2 + 2n + 1 which is NOT a perfect square!

Answer 3

1. I am going to assume you wanted to write
(n+2)^3 = n^3 + 2^3
Let n=1
(1+2)^3 = 1^3 + 2^3
3^3 = 1^3 +2^3
27=1+8
27=9 not true

2. x^2>x
Let x=0
0^2>0
0>0 not true

If there was supposed to be a >= sign, try with x=.1
.1^2>.1
.01>.1 not true

3. n^2 + n +11 does not factor. As far as I can tell this (tried it for the first six numbers) is true.

4. This also appears to be true for the first six cases. Actually, this appears to be a very interesting fact for perfect squares.
ETA: If you need to prove it:
The sum of the first n integers is
S = n(n+1)/2.

Multiply by 2 to get for only even numbers
S=n(n+1)
So the equation is
n(n+1) *4 +1
4n^2 +4n+1
(2n+1)(2n+1)
(2n+1)^2

This is always true.