標題:
數列求Sn通項公式
發問:
S1=3,S2=7,S3=13,S4=23,S5=41....求Sn... 諗左好耐都諗唔到....希望有人幫到我=[
S1 = 3 = (2 + 2^0)S2 = 7 = (2 + 2^0) + (2 + 2^1)S3 = 13 = (2 + 2^0) + (2 + 2^1) + (2 + 2^2)S4 = 23 = (2 + 2^0) + (2 + 2^1) + (2 + 2^2) + (2 + 2^3) S5 = 41 = (2 + 2^0) + (2 + 2^1) + (2 + 2^2) + (2 + 2^3) + (2 + 2^4)..........................................................................................................S(n) = (2 + 2^0) + (2 + 2^1) + (2 + 2^2) + (2 + 2^3) + ... + (2 + 2^(n-1))S(n) = (2 + 2 + 2 + 2 + ... + 2) + (2^0 + 2^1 + 2^2 + 2^3 + ... + 2^(n-1))S(n) = 2n + (2^0 + 2^1 + 2^2 + 2^3 + ... + 2^(n-1))S(n) = 2n + (2^0)(1 - 2^n) / (1 - 2)S(n) = 2n + (2^n - 1)
其他解答:
3,7,13,23,41,...... 4,6,10,18,...... 2,4,8,......
數列求Sn通項公式
發問:
S1=3,S2=7,S3=13,S4=23,S5=41....求Sn... 諗左好耐都諗唔到....希望有人幫到我=[
此文章來自奇摩知識+如有不便請留言告知
最佳解答:S1 = 3 = (2 + 2^0)S2 = 7 = (2 + 2^0) + (2 + 2^1)S3 = 13 = (2 + 2^0) + (2 + 2^1) + (2 + 2^2)S4 = 23 = (2 + 2^0) + (2 + 2^1) + (2 + 2^2) + (2 + 2^3) S5 = 41 = (2 + 2^0) + (2 + 2^1) + (2 + 2^2) + (2 + 2^3) + (2 + 2^4)..........................................................................................................S(n) = (2 + 2^0) + (2 + 2^1) + (2 + 2^2) + (2 + 2^3) + ... + (2 + 2^(n-1))S(n) = (2 + 2 + 2 + 2 + ... + 2) + (2^0 + 2^1 + 2^2 + 2^3 + ... + 2^(n-1))S(n) = 2n + (2^0 + 2^1 + 2^2 + 2^3 + ... + 2^(n-1))S(n) = 2n + (2^0)(1 - 2^n) / (1 - 2)S(n) = 2n + (2^n - 1)
其他解答:
3,7,13,23,41,...... 4,6,10,18,...... 2,4,8,......
文章標籤
全站熱搜
留言列表