1/2)设Sn是正项数列【an】的前n项和,4Sn=an^2+2an-3.已知bn=2^n,求Tn=a1b1+a2b2+...+anbn的值
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![1/2)设Sn是正项数列【an】的前n项和,4Sn=an^2+2an-3.已知bn=2^n,求Tn=a1b1+a2b2+...+anbn的值](/uploads/image/z/5438407-31-7.jpg?t=1%2F2%29%E8%AE%BESn%E6%98%AF%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%E3%80%90an%E3%80%91%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C4Sn%3Dan%5E2%2B2an-3.%E5%B7%B2%E7%9F%A5bn%3D2%5En%2C%E6%B1%82Tn%3Da1b1%2Ba2b2%2B...%2Banbn%E7%9A%84%E5%80%BC)
1/2)设Sn是正项数列【an】的前n项和,4Sn=an^2+2an-3.已知bn=2^n,求Tn=a1b1+a2b2+...+anbn的值
1/2)设Sn是正项数列【an】的前n项和,4Sn=an^2+2an-3.已知bn=2^n,求Tn=a1b1+a2b2+...+anbn的值
1/2)设Sn是正项数列【an】的前n项和,4Sn=an^2+2an-3.已知bn=2^n,求Tn=a1b1+a2b2+...+anbn的值
不能肯定过程计算正确,但思路应该是对的!
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