Proof of the sum formula of the first n terms of the equal ratio sequence When q = 1 Suppose A1 = 1, n = 2 Sum of the first n terms = 1 + 2 = 3 So the formula Sn = A1 * n is wrong I read fast, and mixed the equal difference with the equal ratio,

Proof of the sum formula of the first n terms of the equal ratio sequence When q = 1 Suppose A1 = 1, n = 2 Sum of the first n terms = 1 + 2 = 3 So the formula Sn = A1 * n is wrong I read fast, and mixed the equal difference with the equal ratio,

Well Lord, you How can you add it like that? If the common ratio is 1, that is to say, the equal ratio sequence is the sequence of equal items, that is, its sum is n'a1. You can't use the two summation formulas in books. Here you say A1 is 1, so when n is 2, then A2 is also 1 So you can't have 1, 2 = 3, so your formula is good at last. You just made a mistake about the hidden condition that every item is 1
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What does 0 + 3 and 1 / 8 - (- 3 / 4 / 5) equal

3

What is 3.75 - (- 1 / 2) + (- 4 2 / 3) + (- 0.5) + (- 6 3 / 4)

3.75-(-1/2)+(-4 2/3)+(-0.5)+(-6 3/4)
＝15/4+1/2-14/3-1/2-27/4
=-3-14/3
=-23/3