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Sn=1*1/2+3*(1/2)³+5*(1/2)5(5次方的意思)+.....+(2n-1)*(1/2)2n-1(2n-1次方的意思)用错位相减法的方法算出通...
Sn=1*1/2+3*(1/2)³+5*(1/2)5(5次方的意思)+.....+(2n-1)*(1/2)2n-1(2n-1次方的意思)
用错位相减法的方法算出通项 展开
用错位相减法的方法算出通项 展开
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解答:
Sn = 1×½ + 3×(½)³ + 5×(½)^5 + .....+ (2n-1)×(½)^(2n-1)
4Sn = 2 + 3×(½)¹ + 5×(½)³ + .....+(2n-1)×(½)^(2n-3)
4Sn - Sn = 2 + 2×(½) + 2×(½)³ + .....+ 2×(½)^(2n-3) - (2n-1)×(½)^(2n-1)
3Sn = 2 + 2×[½ + (½)³ + .....+ (½)^(2n-3)] - (2n-1)×(½)^(2n-1)
= 2 + 2×{½[1-(¼)^(2n-3)]/(1-¾)} - (2n-1)×(½)^(2n-1)
= 2 + 2[1-(¼)^(2n-3)]/¼ - (2n-1)×(½)^(2n-1)
= 10 - 8×(¼)^(2n-3) - (2n-1)×(½)^(2n-1)
∴Sn =[10 - 8×(¼)^(2n-3) - (2n-1)×(½)^(2n-1)]/3
Sn = 1×½ + 3×(½)³ + 5×(½)^5 + .....+ (2n-1)×(½)^(2n-1)
4Sn = 2 + 3×(½)¹ + 5×(½)³ + .....+(2n-1)×(½)^(2n-3)
4Sn - Sn = 2 + 2×(½) + 2×(½)³ + .....+ 2×(½)^(2n-3) - (2n-1)×(½)^(2n-1)
3Sn = 2 + 2×[½ + (½)³ + .....+ (½)^(2n-3)] - (2n-1)×(½)^(2n-1)
= 2 + 2×{½[1-(¼)^(2n-3)]/(1-¾)} - (2n-1)×(½)^(2n-1)
= 2 + 2[1-(¼)^(2n-3)]/¼ - (2n-1)×(½)^(2n-1)
= 10 - 8×(¼)^(2n-3) - (2n-1)×(½)^(2n-1)
∴Sn =[10 - 8×(¼)^(2n-3) - (2n-1)×(½)^(2n-1)]/3
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不懂,很深奥,你把分给我吧,反正你浪费也是浪费
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(1/2)^2*Sn=1*(1/2)³+3*(1/2)^5+5*(1/2)^7+.....+(2n-1)*(1/2)^(2n+1),
(1-(1/2)^2)Sn=1*1/2-2*(1/2)³-2*(1/2)^5-……-2*)*(1/2)^(2n+1)
再用等比数列求和算
(1-(1/2)^2)Sn=1*1/2-2*(1/2)³-2*(1/2)^5-……-2*)*(1/2)^(2n+1)
再用等比数列求和算
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Sn=1*(1/2)+3*(1/2)^3+5*(1/2)^5+.....+(2n-1)*(1/2)^(2n-1),
(1/4)*Sn=1*(1/2)+3*(1/2)^5+5*(1/2)^7+.....+(2n-1)*(1/2)^(2n+1),
所以,上等式-下等式为
(3/4)*Sn=(1/2)+(1/2)^2+(1/2)^4+......+(1/2)^(2n-2)-(2n-1)*(1/2)^(2n+1),
即(3/4)*Sn=(1/2)+(1/3)*[1-(1/4)^(n-1)]-(2n-1)*(1/2)^(2n+1),
整理得:Sn=(4/3)*{(1/2)+(1/3)*[1-(1/4)^(n-1)]-(2n-1)*(1/2)^(2n+1)}
(1/4)*Sn=1*(1/2)+3*(1/2)^5+5*(1/2)^7+.....+(2n-1)*(1/2)^(2n+1),
所以,上等式-下等式为
(3/4)*Sn=(1/2)+(1/2)^2+(1/2)^4+......+(1/2)^(2n-2)-(2n-1)*(1/2)^(2n+1),
即(3/4)*Sn=(1/2)+(1/3)*[1-(1/4)^(n-1)]-(2n-1)*(1/2)^(2n+1),
整理得:Sn=(4/3)*{(1/2)+(1/3)*[1-(1/4)^(n-1)]-(2n-1)*(1/2)^(2n+1)}
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