Q&A - P3 Math

[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.

[Dharma]

ks2me:

Dharma:

[quote=\"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.[/quote]Those in P4 doing Olympiad Maths will be able to appreciate.

Show them square numbers (series with consecutive odd nos) and triangular numbers in pictorial form. Will understand better after watching this http://www.yteach.co.uk/index.php/resources/triangular_square_numbers_root_t_page_3.html

[Muffins]

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 for the crystal clear explaining, buddy! :thankyou: :salute:

[small]

[quote]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[/quote]Hi Dharma,

Thanks for sharing...learned something new... :celebrate: :udaman: