What is the sum of the first 150 odd positive integers?

There are many approaches. Here's one...
Hint: The sum of the odd integers appear on a diagonal of the multiplication table.
 
Try it for the first few:

First 1 odd: 1

First 2 odd: 1+3 = 4

First 3 odd: 1+3+5 = 9

First 4 odd: 1+3+5+7 = 16

First 5 odd: 1+3+5+7+9 = 25

See a pattern?
 
helping said:
What is the sum of the first 150 odd positive integers?
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There is a formula for the first 300 integers (150 odd, 150 even) do you know that formula?
The sum of the first 150 even integers is 2+ 4+ 6+ ...+ 298+ 300= 2(1+ 2+ 3+ ...+ 150) so find the sum of the first 300 positive integers and subtract 2 times the sum of the first 150 integers.
 
daon2 said:
Try it for the first few:

First 1 odd: 1

First 2 odd: 1+3 = 4

First 3 odd: 1+3+5 = 9

First 4 odd: 1+3+5+7 = 16

First 5 odd: 1+3+5+7+9 = 25

See a pattern?
Click to expand...

Thanks,

Yes, they are perfect squares of the first 5 numbers, meaning I can just square 150 and the answer is 22,500.

How do I explain it to my daughter though. Not that you have to square 150 or how to do it, but that the first thing she should have done is write the first few like you have, then pay attention to the answers,

What made you think to write it like you have in the first place, absent your knowledge of it beforehand?
 
helping said:
How do I explain it to my daughter
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In mathematics, it's called "taking cases".

In the real world, it's called "looking for patterns".

Don't be afraid to do something. You can't break anything by experimenting.

Food for thought.
 
To add to the above, it really is standard practice. Another name for this kind of problem might be "find a formula for _________." When trying to show something is true, or discover truths, many examples should be examined, and that is when the "pattern searching" begins (If you had only seen two examples above, it would be hard to make an accurate guess). This problem is popularly (re-)introduced to Calculus students with this story of Gauss.

http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss said:
Another famous story has it that in primary school after the young Gauss misbehaved, his teacher, J.G. Büttner, gave him a task : add a list of integers inarithmetic progression; as the story is most often told, these were the numbers from 1 to 100. The young Gauss reputedly produced the correct answer within seconds, to the astonishment of his teacher and his assistant Martin Bartels.
Gauss's presumed method was to realize that pairwise addition of terms from opposite ends of the list yielded identical intermediate sums: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on, for a total sum of 50 × 101 = 5050. However, the details of the story are at best uncertain (see[26] for discussion of the original Wolfgang Sartorius von Waltershausen source and the changes in other versions); some authors, such as Joseph Rotman in his book A first course in Abstract Algebra, question whether it ever happened.
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You can also use Gauss' technique to solve your daughter's problem. Try it!
 
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To add to what daon and mmm have said: when faced with a problem that you do not know how to solve, it is a common technique, in mathematics and other fields, to try simpler examples of the same type of problem in the hope of finding a solution or a pattern. It is so common that it has a name: "induction."

Mathematicians do not view such induction as proof, but induction does suggest to them what to prove and sometimes how to prove it. They then have greatly confused the issue by labelling one of their accepted methods of proof "mathematical induction," which is usually not taught until college math. I would not expect someone studying beginning algebra to do anything more than find a pattern by induction
 
How do I explain it to my daughter though.
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First:
Define nth odd number.

odd:1,3,5,7,...,2n-1
n:1,2,3,4,...,nth


Second:
Re-state the problem. Define SUM(n).

Find the SUM(n) of the first n odd numbers {1, 3, ..., 2n-1}
SUM(n) = 1 + 3 + ... + 2n-1


Third:
Derive formula for SUM(n)

SUM(n)=1+3+5+...+2n-5+2n-3+2n-1
SUM(n)=2n-1+2n-3+2n-5+...+5+3+1
--------
2SUM(n)=2n+2n+2n+...+2n+2n+2n

Notice that we get n identical terms of 2n

2SUM(n)= n*2n
2SUM(n)= 2n2
SUM(n)= n2
 
If your daughter is given a problem like this, then she probably knows, and can use, the formula for the sum of the first "n" numbers. That's one of the difficulties with you posting rather than her- we cannot ask directly.

If you or she does not know that formula, write out
1+ 2 + 3+ ...+ n-1+ n and reverse it
n+ n-1+ n-2+ ..._ 2 + 1 and add each pair vertically.

The sum of each pair of numbers is n+ 1 and there are n pairs so the sum of both sets of numbers is n(n+1). Finally, because there are two sums from 1 to n there, divide by 2: the sum of all integers from 1 to n is n(n+1)/2. The sum of all integers from 1 to 300 is (300(301)/2= 150(301)= 45150

we can find a formula for the sum of even numbers from 2 to 2n by writing 2+ 4+ 6 +...+ 2n-2+ 2n= 2(1)+ 2(2)+ 2(3)+ ...+ 2(n-1)+ 2n= 2(1+ 2+ 3+ ...+ n-1+ n)= 2(n(n+1)/2)= n(n+ 1). The sum of even integers from 1 to 300= 2(150) is 150(151)= 22650.

And since all numbers are either even or odd, we can find the sum of the first 150 odd integers, that is the sum of all odd integers from 1 to 299 by subtracting the sum of all even numbers from 2 to 300 from the set of all integers from 1 to 300: 45150- 22650= 22500.
 
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