Find the trick and go for it...
Find the value of
1 ≤ a < b < c ∑ 2 a 3 b 5 c 1
That is the sum of 2 a 3 b 5 c 1 over all triples of positive integers ( a , b , c ) satisfying a < b < c .
Give your answers to 4 decimal places.
The answer is 0.0006.
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2 solutions
S = 1 ≤ a < b < c ∑ 2 a 3 b 5 c 1 = a = 1 ∑ ∞ 2 a 1 b = a + 1 ∑ ∞ 3 b 1 c = b + 1 ∑ ∞ 5 c 1 = a = 1 ∑ ∞ 2 a 1 b = a + 1 ∑ ∞ 3 b 1 ⋅ 5 b + 1 1 c = 0 ∑ ∞ 5 c 1 = a = 1 ∑ ∞ 2 a 1 b = a + 1 ∑ ∞ 3 b 1 ⋅ 5 b + 1 1 ⋅ 4 5 = 4 1 a = 1 ∑ ∞ 2 a 1 b = a + 1 ∑ ∞ 1 5 b 1 = 4 1 ⋅ 1 4 1 a = 1 ∑ ∞ 2 a 1 ⋅ 1 5 a 1 = 4 1 ⋅ 1 4 1 a = 1 ∑ ∞ 3 0 a 1 = 4 1 ⋅ 1 4 1 ⋅ 2 9 1 ≈ 0 . 0 0 0 6 Note that k = 0 ∑ ∞ ( a 1 ) k = 1 − a 1 1 = a − 1 a
But in the question we can consider that as a GP series. In that case the answer is 0.000459770115. Why is it different? Please explain.
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How did you get 0.000459770115?
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Here r=1/{(2) (3) (5)} And a=1/{(2) (3^2) (5^3)} Then we will apply summation of GP of infinite series, as r<1.
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@Bivab Mishra – Sorry sorry you are correct ,I considered only one case.
@Bivab Mishra – You have included 2 2 ⋅ 3 2 ⋅ 5 3 1 , 2 ⋅ 3 3 ⋅ 5 3 1 , 2 4 ⋅ 5 ⋅ 5 3 1 and many other terms that are not to be included.
For positive integers a , x , y , we sum ∑ 2 a 1 3 a + x 1 5 a + x + y 1 = ∑ 3 0 a 1 1 5 x 1 5 y 1 = 2 9 1 1 4 1 4 1 = 1 6 2 4 1 ≈ 0 . 0 0 0 6 1 6