1) i+i^2+i^3+...+i^15
2) (√1 + √2)(√1 - √2) + (√2 + √3)(√2 - √3) + ... +(√n + √n+1)(√n - √n+1)
更新1:
√ = root
f.4 maths about complex number
2009-09-16 7:51 am
plz help me to calculate/simplify the questions below,
1) i+i^2+i^3+...+i^15
2) (√1 + √2)(√1 - √2) + (√2 + √3)(√2 - √3) + ... +(√n + √n+1)(√n - √n+1)
1) i+i^2+i^3+...+i^15
2) (√1 + √2)(√1 - √2) + (√2 + √3)(√2 - √3) + ... +(√n + √n+1)(√n - √n+1)
回答 (2)
2009-09-16 7:55 am
✔ 最佳答案
1) i + i2 + i3 + ... + i15 = (i - 1 - i + 1) repeated 3 times + i - 1 - i
= 0 * 3 - 1
= -1
2) (√1 + √2)(√1 - √2) + (√2 + √3)(√2 - √3) + ... +(√n + √n+1)(√n - √n+1)
= (1 - 2) + (2 - 3) + ... + (n - n - 1) <-- n terms, each term = -1
= (-1) * n
= -n
2009-09-16 8:12 am
1)i+i^2+i^3+...+i^15 is a G.S. of 1st term is i and common ratio also is i
= i (1 - i^15) / (1 - i)
= i (1 + i) / (1 - i)
= (i - 1) / (1 - i)
= - 1
= i (1 - i^15) / (1 - i)
= i (1 + i) / (1 - i)
= (i - 1) / (1 - i)
= - 1
收錄日期: 2021-04-21 22:03:35
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https://hk.answers.yahoo.com/question/index?qid=20090915000051KK02448