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Secondary 1 | Maths
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liang liang
Liang Liang

Secondary 1 chevron_right Maths chevron_right Singapore

Pls help. I need the explanation also .

Date Posted: 2 years ago
Views: 365
AC Lim
AC Lim
2 years ago
For more than 200
15 × 15 = 225
Hence,
1+3+5+7+9+11+13+15+17+19+21+23+25+27+29 = 225
n = 29

Or

15×2 -1 =29
liang liang
Liang Liang
2 years ago
can you explain it more clearly
Eric Nicholas K
Eric Nicholas K
2 years ago
The point is that 1, 4, 9, 16, 25… are perfect squares, because square-rooting these values give us integers.

If we take the square root of 200, we get 14.14. Upon closer inspection, 14^2 gives 196 (^ represents to the power of) while 15^2 gives 225.

We can see by inspection that we need the row which contains … = 225 = 15 x 15.
Eric Nicholas K
Eric Nicholas K
2 years ago
But, which line has the numbers to get us 15 x 15?

Upon inspection, we see that we need the first 15 numbers in the pattern 1 + 3 + 5 + …, and if you list out everything, 15th term will be 29.

1, 3, 5, 7, 9,
11, 13, 15, 17, 19,
21, 23, 25, 27, 29

So, we have the line 1 + 3 + 5 + … + 27 + 29 = 225 = 15 x 15, and therefore n = 29.
J
J
2 years ago
No, it should be that they are perfect squares because they are products of a positive integer and itself. (Hence the term square since the analogy comes from a square's area)

The latter causes the former.


Square rooting the perfect square gives back the integer.


Square rooting to get integer is the reverse of the result, not the cause of it.
J
J
2 years ago
How to present the working :


Based on the pattern, we can deduce that :

1 + 3 + 5 + 7 + 9 +... + n = k² , where is a integer


So when 1 + 3 + 5 + 7 + 9 +... + n > 200 ,

k² > 200

k > √200 (since n is positive, we do not consider the negative square root)

k > 14.14 (4.sf)

Least value of k = 15 since 15 is the smallest integer bigger than it.



Next, from the pattern we can see that there are k odd numbers for the kth row.

(In other words, the number of odd numbers in each row is the same as the row number)


We observe that the last number, n equals to 2k - 1


Eg.

1
3 = 2(2) - 1
5 = 2(3) - 1
7 = 2(4) - 1
9 = 2(5) - 1

So n = 2k - 1


So least value of n

= 2(15) - 1
= 30 - 1
= 29


Alternatively,


k > √200

2k > 2√200

2k - 1 > 2√200 - 1

2k - 1 > 27.28

n > 27.28

Since n is odd, least value of n = 29

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