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1 solution
Hello @Kay do you know where i could learn the theorem of double sum(like the question stated above)?
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Usually I don't have particulars, but I expand the summation. This may be done if it is not complicated, but if it is, then we will have to look for other way. Alternatively for this question, one can simply expand the a summation first and make it to b = 0 ∑ 4 ( 1 b + 2 b + 3 b + 4 b ) = 4 + 1 0 + 3 0 + 1 0 0 + 3 5 4 = 4 9 8
From where u got the formula of eleminating b
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None. You just have to distribute the B sigma into a 0 + a 1 + a 2 + a 3 + a 4 . The rest depends on how you approach this A sigma
The most simple way is the rewrote b summation as 1 + a + a 2 + a 3 + a 4 = a − 1 a 5 − 1 . This formula is valid for a = 1 . As the summation requires a = 1 , then the first term of a summation shall be 5. The rest is 5 + a = 2 ∑ 4 a − 1 a 5 − 1 = 5 + 3 1 + 2 2 4 2 + 3 1 0 2 3 = 4 9 8