Q&A - P3 Math
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Dharma:
Thanks Dharma, interesting pattern observation, learn something new again.
Hi Suz855,ks2me & small,
Hope this may be of interest.
1/2006 is the common factor of the series.
1/2006( 1+3+5+ā¦.+ 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+ā¦.+2003+2005
To find the number of terms (2005+1)/2 = 1003
So, 1/2006( 1+3+5+ā¦+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5
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ks2me:
Hi ks2me
Thanks Dharma, interesting pattern observation, learn something new again.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+ā¦.+ 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+ā¦.+2003+2005
To find the number of terms (2005+1)/2 = 1003
So, 1/2006( 1+3+5+ā¦+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5
This is slight variation to this sort of question.
2/2006 + 4/2006 + 6/2006 + ā¦+ 2004/2006 + 2006/2006
2/2006(1+2+3+ā¦+ 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+ ā¦.+ (N-1)+N =N(N+1)/2
2/2006(1+2+3+ā¦+ 1002+1003)
=2/2006 x 1003x1004/2
= 502 -
Great sharing Dharma! :salute:
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ks2me:
Great sharing Dharma! :salute:
Glad u like it -
Dharma:
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.ks2me:
Great sharing Dharma! :salute:
Glad u like it -
ks2me:
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.Dharma:
[quote=\"ks2me\"]Great sharing Dharma! :salute:
Glad u like it
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 -
Dharma:
Thanks for the crystal clear explaining, buddy! :thankyou: :salute:Hi Suz855,ks2me & small,
Hope this may be of interest.
1/2006 is the common factor of the series.
1/2006( 1+3+5+ā¦.+ 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+ā¦.+2003+2005
To find the number of terms (2005+1)/2 = 1003
So, 1/2006( 1+3+5+ā¦+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5 -
[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+ā¦.+ 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+ā¦.+2003+2005
To find the number of terms (2005+1)/2 = 1003
So, 1/2006( 1+3+5+ā¦+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5[/quote]Hi Dharma,
Thanks for sharing...learned something new... :celebrate: :udaman: -
Pls help me with my son's H/W...
Minghua gave 1/2 of his money to Lily,.
In return, Lily gave 1/3 of whatever money
she had to Minghua.
Later, Minghua gave 1/4 of whatever
money he had then to Lily.
Finally, Minghua had $675 and Lily had $1325.
how much money di Minghua have at first? -
is the answer $700 ?
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