一些數學題20點 謝謝,需詳解

1) 等比數列 0.5+1.5+4.5+.........+1093.5
首項 a = 0.5
公比 R = 1.5/0.5 = 3
假設第n項為1093.5
a*R^(n-1) = 1093.5
0.5*3^(n-1) = 1093.5
3^(n-1) = 1093.5/0.5 = 2187
(n-1)*log 3 = log 2187
n-1 = log 2187 / log 3
n -1 = 7
n = 8
0.5+1.5+4.5+.........+1093.5 = a*(R^n - 1)/(R - 1)
= 0.5*(3^8 - 1) / (3-1)
= 0.5*6560/2
= 1640.

2) 首項 a = 2*1+5 = 7
公差 d = [2(n+1) + 5] - [2n+5]
= 2
首25項的集和 = n/2*[2*a + (n-1)*d]
= 25/2*[2*7 + (25-1)*2]
= 25/2*(62)
= 775

3) 假設 a 為首項, R為公比.
aR² = 6.75......(1)
aR³ = 10.125......(2)
(2)/(1): R = 10.125/6.75 = 1.5
a(1.5)² = 6.75
=> a = 3
等比級數第一至第四項的和
= a*(R^(n-1)-1)/(R-1)
= 3*[1.5^3-1)/(1.5-1)
= 14.25

4) 假設R為公比.
等比級數首項是24及級數的無限項之和是36
=> 24/(1-R) = 36
1-R = 24/36 = 2/3
R = 1/3.
所以公比是 1/3.

1)
0.5+1.5+4.5+...+1093.5
=(1+3+9+...+2187)/2
=(2+6+18+...+4374)/(2*2)
=(6561-1)/4
=6560/4
=3280/2
=1640
2)
[2(1)+5]+[2(2)+5]+...+[2(25)+5]
=25{[2(1)+5]+[2(25)+5]}/2
=25(7+55)/2
=25(31)
=775
3)
T(n)/T(n-1)=10.125/6.75=1.5
T(1)=T(3)/(1.5)^2=6.75/2.25=3
T(2)=3*1.5=4.5
S(4)
=3+4.5+6.75+10.125
=24.375
4)
T(1)/24=1
S(Unlimit)/24=1.5
=1/(1-r) (r---公比)
1-r=2/3
-r=-1/3
r=1/3
Therefore,公比=1/3.

1) 等比數列 0.5+1.5+4.5+.........+1093.5
首項 a = 0.5
公比 R = 1.5/0.5 = 3
假設第n項為1093.5
a*R^(n-1) = 1093.5
0.5*3^(n-1) = 1093.5
3^(n-1) = 1093.5/0.5 = 2187
(n-1)*log 3 = log 2187
n-1 = log 2187 / log 3
n -1 = 7
n = 8
0.5+1.5+4.5+.........+1093.5 = a*(R^n - 1)/(R - 1)
= 0.5*(3^8 - 1) / (3-1)
= 0.5*6560/2
= 1640.

2) 首項 a = 2*1+5 = 7
公差 d = [2(n+1) + 5] - [2n+5]
= 2
首25項的集和 = n/2*[2*a + (n-1)*d]
= 25/2*[2*7 + (25-1)*2]
= 25/2*(62)
= 775

3) 假設 a 為首項, R為公比.
aR² = 6.75......(1)
aR³ = 10.125......(2)
(2)/(1): R = 10.125/6.75 = 1.5
a(1.5)² = 6.75
=> a = 3
等比級數第一至第四項的和
= a*(R^(n-1)-1)/(R-1)
= 3*[1.5^3-1)/(1.5-1)
= 14.25

4) 假設R為公比.
等比級數首項是24及級數的無限項之和是36
=> 24/(1-R) = 36
1-R = 24/36 = 2/3
R = 1/3.
所以公比是 1/3.

I don't know,sorry about that