Maths GCSE help me pls

hiii please can someone help me with proof questions? like if you got asked to prove that the sum of any 3 consecutive odd no.s is always 11 more that a multiple of 12?? I get how to work put all the equation shiz I just don't know what to write to prove it, thankss

Reply 1

xoxbeckywxox

6

Are you sure it's not sum of squares of 3 consecutive odd numbers?

we are looking for 12M +11 to come out of the proof (where M is any expression)
3 consecutive odd numbers:
2n+1, 2n+3, 2n+5

square them:
4n^2+4n+1, 4n^2+12n+9, 4n^2+20n+25

Add them:
12n^2+36n+35

take 12 outside the brackets:
12(n^2+3n+2) +11 (because 2x12 is 24, then you add 11 to get 35)

since anything multiplied by 12 is a multiple of 12, we can say this proves the statement.

so write: M= n^2+3n+2.
12(n^2+3n+2) +11 = 12M +11, where M is an integer. Therefore, the sum of squares of three consecutive whole numbers is 11 more than a multiple of 12.

Hopefully that helps a bit.

(edited 6 years ago)

Reply 2

3317752

13

Original post by Cait157

hiii please can someone help me with proof questions? like if you got asked to prove that the sum of any 3 consecutive odd no.s is always 11 more that a multiple of 12?? I get how to work put all the equation shiz I just don't know what to write to prove it, thankss

imagine as if ur explaining to a person in person. Write down everything u would tell that person along with the equations. Alternately, just write the equations down and solve it. The examiner will look at ur working out and he/she will find out.

Reply 3

Cait157

OP

12

Original post by xoxbeckywxox

Are you sure it's not sum of squares of 3 consecutive odd numbers?

we are looking for 12M +11 to come out of the proof (where M is any expression)
3 consecutive odd numbers:
2n+1, 2n+3, 2n+5

square them:
4n^2+4n+1, 4n^2+12n+9, 4n^2+20n+25

Add them:
12n^2+36n+35

take 12 outside the brackets:
12(n^2+3n+2) +11 (because 2x12 is 24, then you add 11 to get 35)

since anything multiplied by 12 is a multiple of 12, we can say this proves the statement.

so write: M= n^2+3n+2.
12(n^2+3n+2) +11 = 12M +11, where M is an integer. Therefore, the sum of squares of three consecutive whole numbers is 11 more than a multiple of 12.

Hopefully that helps a bit.

ohhh I never took the 12 out of the brackets, thank you!!