Q&A - P3 Math

[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

[corneyAmber]

Dharma:

ks2me:

Great sharing Dharma! :salute:

Glad u like it

Yes I do but I wonder if P3/P4 children can appreciate it? My child is not at that level yet so cannot test on her.