請選擇任何一條等式證明﹕
A. 1+2+4+8+…+2n=2(2n)-1
B. 1+2+4+8+…+2n=2n+1-1
最佳解答者得10分﹕數學證明題
2007-03-06 3:18 am
回答 (4)
2007-03-06 3:43 am
2007-03-06 7:57 am
因為1+(1+2+4+8+...+2^n)
=2+2+4+8+...+2^n
=4+4+8+...+2^n
=....
=2^n+2^n
=2(2^n)
所以1+2+4+8+...+2^n=2(2^n)-1=2^(n+1)-1(A.B.同時得證)
=2+2+4+8+...+2^n
=4+4+8+...+2^n
=....
=2^n+2^n
=2(2^n)
所以1+2+4+8+...+2^n=2(2^n)-1=2^(n+1)-1(A.B.同時得證)
2007-03-06 3:38 am
A.
it is a geometric sequence
1 + 2 + 4 + 8 + .... + 2^n
a = 1
R = 2
T(k) = aR^(k - 1)
2n = 2^(k - 1)
n + 1 = k
so,there is n + 1 terms
by sum of geometric sequence
LHS
= 1 + 2 + 4 + 8 + ..... + 2^n
= a(R^k - 1) / (R - 1)
= 2^(n + 1) - 1
= 2(2^n) - 1
= RHS
B.
LHS
= 1+2+4+8+…+2n
= 2(2^n) - 1(proved)
= 2^(n + 1) - 1
= RHS
it is a geometric sequence
1 + 2 + 4 + 8 + .... + 2^n
a = 1
R = 2
T(k) = aR^(k - 1)
2n = 2^(k - 1)
n + 1 = k
so,there is n + 1 terms
by sum of geometric sequence
LHS
= 1 + 2 + 4 + 8 + ..... + 2^n
= a(R^k - 1) / (R - 1)
= 2^(n + 1) - 1
= 2(2^n) - 1
= RHS
B.
LHS
= 1+2+4+8+…+2n
= 2(2^n) - 1(proved)
= 2^(n + 1) - 1
= RHS
參考: eason mensa
2007-03-06 3:22 am
B. 1+2+4+8+…+2n=2n+1-1
收錄日期: 2021-04-23 00:10:01
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