O-Level Additional Math
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Hi, need help on another Q. Kindly help. Thank you.
Evaluate 2012^2 - 2011^2 + 2010^2 - 2009^2 + ...
+ 4^2 - 3^2 + 2^2 - 1^2
:? :?:
:thankyou: -
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nounou:
Use this formulae : a^2 ā b^2 = (a + b)(a ā b)Hi, need help on another Q. Kindly help. Thank you.
Evaluate (2012^2 - 2011^2) + (2010^2 - 2009^2) + ...
+ (4^2 - 3^2) + (2^2 - 1^2)
:? :?:
:thankyou:
2012^2 - 2011^2
= (2012+2011)(2012-2011)
= 2012+2011
2010^2 - 2009^2
= (2010+2009)(2010-2009)
= 2010+2009
4^2 - 3^2
= (4+3)(4-3)
= 4+3
2^2 - 1^2
= (2+1)(2-1)
= (2+1)
so you need to add 1+2+3+4 .... +2009+2010+2011+2012 to get the ans.
hopefully someone else can give a shorter working answer. -
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Hi, thank you very much for your help. I understand your workings.
Q: how to find 1+2+3+4 .... +2009+2010+2011+2012 ???? :?:
Please help. :thankyou: :lovesite:koguma:
Use this formulae : a^2 ā b^2 = (a + b)(a ā b)nounou:
Hi, need help on another Q. Kindly help. Thank you.
Evaluate (2012^2 - 2011^2) + (2010^2 - 2009^2) + ...
+ (4^2 - 3^2) + (2^2 - 1^2)
:? :?:
:thankyou:
2012^2 - 2011^2
= (2012+2011)(2012-2011)
= 2012+2011
2010^2 - 2009^2
= (2010+2009)(2010-2009)
= 2010+2009
4^2 - 3^2
= (4+3)(4-3)
= 4+3
2^2 - 1^2
= (2+1)(2-1)
= (2+1)
so you need to add 1+2+3+4 .... +2009+2010+2011+2012 to get the ans.
hopefully someone else can give a shorter working answer. -
Hi nounou,
To solve for 1+2+3+ā¦+2010+2011+2012, you can group them together as follow:
(1+2012)+(2+2011)+(3+2010)+ā¦
There will be a total of 2012/2=1006 pairs.
Hence 1006 x 2013 = 2,025,078.
Hope it helps.
Cheers,
Jtutor -
:thankyou: :udaman: :lovesite:
Jtutor:
Hi nounou,
To solve for 1+2+3+..............+2010+2011+2012, you can group them together as follow:
(1+2012)+(2+2011)+(3+2010)+...
There will be a total of 2012/2=1006 pairs.
Hence 1006 x 2013 = 2,025,078.
Hope it helps.
Cheers,
Jtutor -
Can anyone recommend physics tution arround bukit timah?
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Hi
Please help.
Secondary Two Maths.
Factorise
1)2r^2-5r-3
Expand and simplify the following expressions
1)(a+b)^2 - (a-b)^2
2)(x+2y)(x-6y)- (x-3y)(x-y)
TIA
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