作业帮短时间里一定采纳,)设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.(1)求a3,a4;(1)当n=1时,a1=S1=2a1-2,解得a1=S1=2.由S[n]=2a[n]-2^n 得 S[n+1]=2a[n+1]-2^(n+1),两式相减并整理得a[n+1] = 2a[
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![短时间里一定采纳,)设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.(1)求a3,a4;(1)当n=1时,a1=S1=2a1-2,解得a1=S1=2.由S[n]=2a[n]-2^n 得 S[n+1]=2a[n+1]-2^(n+1),两式相减并整理得a[n+1] = 2a[](/uploads/image/z/1999771-43-1.jpg?t=%E7%9F%AD%E6%97%B6%E9%97%B4%E9%87%8C%E4%B8%80%E5%AE%9A%E9%87%87%E7%BA%B3%2C%29%E8%AE%BE%E6%95%B0%E5%88%97%7Ba%5Bn%5D%7D%E7%9A%84%E5%89%8D+n+%E9%A1%B9%E5%92%8C+S%5Bn%5D%3D2a%5Bn%5D-2%5En.%E8%AE%BE%E6%95%B0%E5%88%97%7Ba%5Bn%5D%7D%E7%9A%84%E5%89%8D+n+%E9%A1%B9%E5%92%8C+S%5Bn%5D%3D2a%5Bn%5D-2%5En.%281%29%E6%B1%82a3%2Ca4%3B%281%29%E5%BD%93n%3D1%E6%97%B6%2Ca1%3DS1%3D2a1-2%2C%E8%A7%A3%E5%BE%97a1%3DS1%3D2.%E7%94%B1S%5Bn%5D%3D2a%5Bn%5D-2%5En+%E5%BE%97+S%5Bn%2B1%5D%3D2a%5Bn%2B1%5D-2%5E%28n%2B1%29%2C%E4%B8%A4%E5%BC%8F%E7%9B%B8%E5%87%8F%E5%B9%B6%E6%95%B4%E7%90%86%E5%BE%97a%5Bn%2B1%5D+%3D+2a%5B)
短时间里一定采纳,)设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.(1)求a3,a4;(1)当n=1时,a1=S1=2a1-2,解得a1=S1=2.由S[n]=2a[n]-2^n 得 S[n+1]=2a[n+1]-2^(n+1),两式相减并整理得a[n+1] = 2a[
短时间里一定采纳,)设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.
设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.
(1)求a3,a4;
(1)当n=1时,a1=S1=2a1-2,解得a1=S1=2.
由S[n]=2a[n]-2^n 得 S[n+1]=2a[n+1]-2^(n+1),
两式相减并整理得
a[n+1] = 2a[n]+2^n …… (1)
请问 两式相减整理怎么整理到那步?
需要什么公式?..
短时间里一定采纳,ps:"[ ] " 里面的代表比小写还要小的数字,
短时间里一定采纳,)设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.设数列{a[n]}的前 n 项和 S[n]=2a[n]-2^n.(1)求a3,a4;(1)当n=1时,a1=S1=2a1-2,解得a1=S1=2.由S[n]=2a[n]-2^n 得 S[n+1]=2a[n+1]-2^(n+1),两式相减并整理得a[n+1] = 2a[
S[n+1]-S[n]=2a[n+1]-2^(n+1)-2a[n]+2^n
左边：S[n+1]-S[n]=a[n+1] 因为 n+1项和比前n项和多一个 a[n+1]
a[n+1]=2a[n+1]-2a[n]-2^n
得到：a[n+1] = 2a[n]+2^n …… (1) 2^(n+1)-2^n=2*2^n-2^n=2^n