作业帮等比数列a（n)通项公式为a(n)=2ˇn,求算a(i)×a(j)的和T(n) (1≤ i≤ j≤ n) 即 i 和j 为1到n项内任意两项T(n)=a1×[a1＋a2＋a3＋a4＋a5.＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋.a(n-1)×[a(n-1)+a(n)
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![等比数列a（n)通项公式为a(n)=2ˇn,求算a(i)×a(j)的和T(n) (1≤ i≤ j≤ n) 即 i 和j 为1到n项内任意两项T(n)=a1×[a1＋a2＋a3＋a4＋a5.＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋.a(n-1)×[a(n-1)+a(n)](/uploads/image/z/15145674-42-4.jpg?t=%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97a%EF%BC%88n%29%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E4%B8%BAa%28n%29%3D2%CB%87n%2C%E6%B1%82%E7%AE%97a%28i%29%C3%97a%28j%29%E7%9A%84%E5%92%8CT%28n%29+%281%E2%89%A4+i%E2%89%A4+j%E2%89%A4+n%29+%E5%8D%B3+i+%E5%92%8Cj+%E4%B8%BA1%E5%88%B0n%E9%A1%B9%E5%86%85%E4%BB%BB%E6%84%8F%E4%B8%A4%E9%A1%B9T%28n%29%3Da1%C3%97%5Ba1%EF%BC%8Ba2%EF%BC%8Ba3%EF%BC%8Ba4%EF%BC%8Ba5.%EF%BC%8Ba%28n%29%5D%2Ba2%C3%97%5Ba2%EF%BC%8Ba3%EF%BC%8Ba4%EF%BC%8Ba5...%EF%BC%8Ba%28n%29%5D%2Ba3%C3%97%5Ba3%2Ba4%2Ba5...%EF%BC%8Ba%28n%29%5D%EF%BC%8B.a%28n-1%29%C3%97%5Ba%28n-1%29%2Ba%28n%29)
等比数列a（n)通项公式为a(n)=2ˇn,求算a(i)×a(j)的和T(n) (1≤ i≤ j≤ n) 即 i 和j 为1到n项内任意两项T(n)=a1×[a1＋a2＋a3＋a4＋a5.＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋.a(n-1)×[a(n-1)+a(n)
等比数列a（n)通项公式为a(n)=2ˇn,求算a(i)×a(j)的和T(n) (1≤ i≤ j≤ n) 即 i 和j 为1到n项内任意两项
T(n)=a1×[a1＋a2＋a3＋a4＋a5.＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋.a(n-1)×[a(n-1)+a(n)]＋＋a(n)×[a(n)+a(n)]
T(n)=a1×[a1＋a2＋a3＋a4＋a5......＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋..................a(n-1)×[a(n-1)+a(n)]＋＋a(n)×[a(n)+a(n)]这个式子不知道符不符合题意。你们可以自己根据题意列。
等比数列a（n)通项公式为a(n)=2ˇn,求算a(i)×a(j)的和T(n) (1≤ i≤ j≤ n) 即 i 和j 为1到n项内任意两项T(n)=a1×[a1＋a2＋a3＋a4＋a5.＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋.a(n-1)×[a(n-1)+a(n)
∵
a1×[a1＋a2＋a3＋a4＋a5.＋a(n)]
=2×[2×﹙2^n－1﹚÷﹙2－1﹚]
=2²×﹙2^n－1﹚
=2^﹙n＋2﹚－2²
a2×[a2＋a3＋a4＋a5...＋a(n)]
=2²×【2²×[2^﹙n－1﹚－1]÷﹙2－1﹚】
=2⁴×[2^﹙n－1﹚－1]
=2^﹙n＋3﹚－2⁴
a3×[a3+a4+a5...＋a(n)]
=2³×【2³×[2^﹙n－2﹚－1]÷﹙2－1﹚】
=2^﹙n＋4﹚－2^6
……
a(n-1)×[a(n-1)+a(n)]
=2^﹙n－1﹚×【2^﹙n－1﹚×[2^2－1]÷﹙2－1﹚】
=2^﹙2n﹚－2^﹙2n－2﹚
a(n)×[a(n)+a(n)]
=2^n×【2^n×2】
=2^﹙2n＋1﹚
∴T(n)=a1×[a1＋a2＋a3＋a4＋a5.＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋.a(n-1)×[a(n-1)+a(n)]＋＋a(n)×[a(n)+a(n)]
=2^﹙n＋2﹚－2²＋2^﹙n＋3﹚－2⁴＋2^﹙n＋4﹚－2^6＋……＋2^﹙2n﹚－2^﹙2n－2﹚＋2^﹙2n＋1﹚
=【2^﹙n＋2﹚＋2^﹙n＋3﹚＋2^﹙n＋4﹚＋……＋2^﹙2n﹚＋2^﹙2n＋1﹚】－【2²＋2⁴＋2^6＋……＋2^﹙2n－2﹚】
=【2^﹙n＋2﹚×﹙2^n－1﹚÷﹙2－1﹚】－【2²×[2^﹙2n－2﹚－1]÷﹙2－1﹚】
=2^﹙2n＋2﹚－2^﹙n＋2﹚－2^﹙2n﹚＋2²
2^(2n+2)-2^(n+2)-[2^(2n+2)-4]/3
a(i)[a(i)+a(i+1)+a(i+2)+...a(n)] =2^i[2^i(2^(n-i+1)-1)/(2-1)] =2^(n+i+1)-2^(2i) ∑(i=1→n)[2^(n+i+1)-2^(2i)] =∑(i=1→n)2^(n+i+1)-∑(i=1→n)2^(2i) =2^(n+1)∑(i=1→n)2^i-∑(i=1→n)4^i =2^(n+1)2(2^n-1)-4(4^n-1)/(4-1) =4*[2^(2n+1)-3*2^n+1]/3
首先求Sn=a1+a2+a3...an=2^(n+1)-2
再求 Yn= a1*a1+a2*a2+...+an*an=1/3*(4^(n+1)-4)
可知 (Tn-Yn)*2+Yn=Sn*Sn
求解可得
Tn=1/3*2^(2n+3)-2^(n+2)+4/3
顺便问一句，根据题意
T(n)=a1×[a1＋a2＋a3＋a4＋a5......＋a...
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首先求Sn=a1+a2+a3...an=2^(n+1)-2
再求 Yn= a1*a1+a2*a2+...+an*an=1/3*(4^(n+1)-4)
可知 (Tn-Yn)*2+Yn=Sn*Sn
求解可得
Tn=1/3*2^(2n+3)-2^(n+2)+4/3
顺便问一句，根据题意
T(n)=a1×[a1＋a2＋a3＋a4＋a5......＋a(n)]+a2×[a2＋a3＋a4＋a5...＋a(n)]+a3×[a3+a4+a5...＋a(n)]＋..................a(n-1)×[a(n-1)+a(n)]＋＋a(n)×[a(n)+a(n)]
的最后一项应该是a(n)×a(n)更好吧。
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