Q&A - P3 Math

[corneyAmber]

small:

Bingo ks2me.


Your method looks more pro.. :celebrate:

Aiyo small, no la, share your image again as it is not visible on screen. No comparison intended, we all just post our learnings here together as parents. Different kids take to different approaches so it is good to share a few methods in order to know what works for the children.

[small]

Hi ks2me.


Sure..I have uploaded the picture again, it is similar to your method.

I used 'U' shade method to solve the question. :oops:

http://www.postimage.org/image.php?v=Pqrd5ES

http://www.postimage.org/image.php?v=Pqrd5ES

[corneyAmber]

Small, your approach is good and concise. :celebrate:

[Suz855]

Thanks KS2me, this qn is taken from an assessment book … as suspect, the answer given was incorrect.


Thanks for your solution, appreciate

[corneyAmber]

Suz855:

Thanks KS2me, this qn is taken from an assessment book ..... as suspect, the answer given was incorrect.


Thanks for your solution, appreciate

You are :welcome:

But geez...P3/P4 children these days having it quite tough hor..

[Dharma]

small:

Suz855:

Need help to validate answer, thanks



http://www.postimage.org/

Hi Suz855,

Hope this helps. My method might not so clear as I am still under learning stage. :oops:

http://s3.postimage.org/rd5ES-fe3f4cfd2c241f4acdbf433ec3730af2.png\">

Hi Suz855,ks2me & small,

Hope this may be of interest.

1/2006 is the common factor of the series.

1/2006( 1+3+5+….+ 2003+2005)

1 \t\t= 1 (1x1)
1+3\t\t= 4 (2x2)
1+3+5\t\t = 9 (3x3)
1+3+5+7 \t= 16(4x4)

To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

Now,
1+3+5+7+….+2003+2005

To find the number of terms (2005+1)/2 = 1003

So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5

[corneyAmber]

Dharma:

Hi Suz855,ks2me & small,

Hope this may be of interest.

1/2006 is the common factor of the series.

1/2006( 1+3+5+….+ 2003+2005)

1 \t\t= 1 (1x1)
1+3\t\t= 4 (2x2)
1+3+5\t\t = 9 (3x3)
1+3+5+7 \t= 16(4x4)

To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

Now,
1+3+5+7+….+2003+2005

To find the number of terms (2005+1)/2 = 1003

So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5

Thanks Dharma, interesting pattern observation, learn something new again.

[Dharma]

ks2me:

Dharma:

Hi Suz855,ks2me & small,

Hope this may be of interest.

1/2006 is the common factor of the series.

1/2006( 1+3+5+….+ 2003+2005)

1 \t\t= 1 (1x1)
1+3\t\t= 4 (2x2)
1+3+5\t\t = 9 (3x3)
1+3+5+7 \t= 16(4x4)

To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

Now,
1+3+5+7+….+2003+2005

To find the number of terms (2005+1)/2 = 1003

So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5

Thanks Dharma, interesting pattern observation, learn something new again.

Hi ks2me

This is slight variation to this sort of question.

2/2006 + 4/2006 + 6/2006 + …+ 2004/2006 + 2006/2006

2/2006(1+2+3+…+ 1002+1003)

Triangular numbers (If you use a dot to represent the sum of every series you will form a triangle)
1 = 1 =(1x2)/2
1+2 = 3 = (3x4)/2
1+2+3 = 6 =(3x4)/2
1+2+3+4 = 10 = (4x5)/2
1+2+3+4+5 = 15 = (5x6)/2
.
.
.
1+2+3+4+ ….+ (N-1)+N =N(N+1)/2

2/2006(1+2+3+…+ 1002+1003)
=2/2006 x 1003x1004/2
= 502

[corneyAmber]

Great sharing Dharma! :salute:

[Dharma]

ks2me:

Great sharing Dharma! :salute:

Glad u like it