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[math] Part (a): Find the sum
a + (a + 1) + (a + 2) + ... + (a + n - 1)
in terms of a and n.
Part (b): Find all pairs of positive integers (a,n) such that n \ge 2 and
a + (a + 1) + (a + 2) + ... + (a + n - 1) = 100.
note: \ge=greater than or equal to.
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Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi lightcreatortwentyjuan
Welcome to the forum.
You should have a line of useful symbols at the top of each forum page which includes this one: ≥
Part (a). You can re-write this as a + a + ... + a (n of these) + 1 + 2 + ... + n-1.
The sum of n-1 natural numbers is given by n(n-1)/2
Part (b)
So you can put the answer for part (a) equal to 100 and look for possible whole numbered solutions.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi bobbym,
Can I have an explanation please?
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Hi;
Looks like you completely ignored post #3 which gives a hint, also gave some info on how you can edit your post to make that ge look like ≥.
Back where I come from all any one ever needs is the answer. We train to back engineer the method (explanation) from the answer. Why? Because we almost always have the answer but never the method, never the purpose. We find the completed product just lying there buried under tons of rock and sand, maybe buried there for thousands of millennia. It is valuable to be able to fill in the blanks when the answer is handed to you.
What were your thoughts on these problems? Please try to solve this now yourself. If you still can not after that I will show what I did.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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