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發問:
1)求等比數列 0.5+1.5+4.5+.........+1093.5 2)求等差數列首25項的集和,其中數列第n項是2n+5 3)等比數列第三及第四項分別是6.75及10.125,求等比級數第四項的和 4)等比級數首項是24及級數的無限項之和是36.求公比
最佳解答:
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.
其他解答:
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
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一些數學題20點 謝謝,需詳解發問:
1)求等比數列 0.5+1.5+4.5+.........+1093.5 2)求等差數列首25項的集和,其中數列第n項是2n+5 3)等比數列第三及第四項分別是6.75及10.125,求等比級數第四項的和 4)等比級數首項是24及級數的無限項之和是36.求公比
最佳解答:
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.
其他解答:
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
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