SOLUTION: 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 greater than 2 and a + (a +
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-> SOLUTION: 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 greater than 2 and a + (a +
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Question 1064992: 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 greater than 2 and a + (a + 1) + (a + 2) + ... + (a + n - 1) = 100. Answer by math_helper(2461) (Show Source):
You can put this solution on YOUR website! Part (a):
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a + (a+1) + (a+2) + … + (a+n-1)
can be re-written:
(a+a+ … + a) + (1+2+ … + n-1)
There are n a's in the first parenthesis, and the 2nd parenthesis sums to
The sum S is:
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Part (b):
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a and n are integers, and both are > 0
We only need to try factors less than or equal to which is approx 14.14, an taking the floor of this value gives an max integer value of 14 to try.
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Factor 200 to see which factors we should try:
200 = 2*2*2*5*5
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We want to try all combinations of the factors given above, such that their product is <= 14.
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Try n=2: 2(2+2a-1) = 200 —> 1+2a = 198 (won't work because 2a = 197 —> a is a non-integer)
Try n=4: 4(4+2a-1) = 200 —> 3+2a = 196 —> 2a = 193 (no good, a is non-integer)
Try n=8: 8(8+2a-1) = 200 —> 7+2a = 25 —> 2a = 18 —> a=9 , satisfy the equality
n=5: 5(5+2a-1) = 200 —> 4+2a = 40 —> 2a = 36 —> a = 18 , satisfy the equality
n=10: 10(10+2a-1) = 200 —> 9+2a = 20 (no good, a is non-integer)
n=20: n is too large
I think that is it.
