(1) First of all, it is easy to get a 1 from the formula of Sn, because S 1=a 1 = 2a2-7 is brought into the formula. Meanwhile, if n=2 is substituted into the formula, S2 = A65438+. Combine two formulas, a 1=3, a2=5, a3=7 because S3= 15, a 1=3, a2=5, a3=7. The above is the standard solution to the first question.
(2) The second problem is the difficulty of this problem. There are many formulas and techniques that can be used to solve the sequence problem. The most common solution to this problem is Sn-Sn- 1=an. Similarly, S(n+ 1)-Sn=a(n+ 1), where n is
Obviously, this formula is not the general formula we need. Next, we use other conditions to observe the first question. According to a 1=3, a2=5, a3=7, it is not difficult for us to guess an=2n+ 1, but guess is guess after all. We need to prove by a more conventional method: mathematics.
We prove it in two cases: ① when n= 1, we substitute the formula in the above figure (name the formula in the figure as formula A) and find that formula A conforms to formula 2n+ 1, that is, we prove that when n= 1, it does satisfy an=2n+ 1.
② We can't just prove that n= 1. We need to prove that when n=k (where k belongs to n*) still conforms to formula A, first assume that n=k conforms, and then prove that n=k+ 1 conforms. Assuming that n = k meets, then an=2k+ 1, then that's it. a(k+ 1)= 2k+3 = 2(k+ 1)+ 1。 Assuming that n=k conforms to formula A, it is proved that n=k+ 1 conforms to formula A, and it is also proved that an=2n+ 1 is general.
The difficult idea used in this problem is that it is necessary to assume that n=k holds, and then prove that n=k+ 1 holds. It can be considered that when 1 is added to this formula continuously, it means that this formula is not only in a certain part, just as we know a 1, a2, a3, then we prove a4 is established, then we know a4 is established.