Prove that (3n+1)²-(3n-1)² is a multiple of 4 taking into account that n is a positive integer value
- Square the brackets (3n+1)²= (3n+1)(3n+1) = 9n²+3n+3n+1 = 9n²+6n+1 (3n-1)²= (3n-1)(3n-1) = 9n²-3n-3n+1 = 9n²-6n+12. Write out the full equation (9n²+6n+1) - (9n²-6n+1) = 9n²+6n+1-9n²+6n-1 = 12n3. Explain your reasoning 12n is divisible by 4 as (12n÷4) equals 3n therefore (3n+1)²-(3n-1)² is a multiple of 4 as 4 goes into 12 a total of 3 times and 3 is an integer
NK
Related Maths GCSE answers
All answers ▸Give an example of a real-world system that would be described by a quadratic equation. Explain the significance of the two real roots, a repeated root, and undefined roots. Is there any significance to a positive or a negative answer in your example?
How do I solve a quadratic equation: x^2-5x+6=0
The range of a set of numbers is 15 1/4. The smallest number is –2 7/8. Work out the largest number.
A linear sequence begins: a + 2b, a + 6b, a + 10b, ..., ... Given that the 2nd term has a value of 8 and the 5th term has a value of 44, calculate the values for a and b