Question

用錯(cuò)位相減法求b_n=n²×2ⁿ的前n項(xiàng)和．

Answer 1

考點(diǎn)：數(shù)列的求和

專題：等差數(shù)列與等比數(shù)列

分析：兩次運(yùn)用錯(cuò)位相減法即可求得bn=n2×2n的前n項(xiàng)和．

解答： 解：由bn=n2×2n，
則其前n項(xiàng)和為Sn=12×21+22×22+32×23+…+n2×2n  ①，
2Sn=12×22+22×23+…+(n-1)2×2n+n2×2n+1  ②，
①-②得：
-Sn=1×21+3×22+5×23+…+(2n-1)×2n-n2×2n+1．
再令Tn=1×21+3×22+5×23+…+(2n-1)×2n  ③，
2Tn=1×22+3×23+…+(2n-3)×2n+(2n-1)×2n+1  ④，
③-④得：-Tn=2+23+24+…+2n+1-(2n-1)×2n+1
=2+

8(1-2n-1)

1-2

-(2n-1)×2n+1=2ⁿ⁺²-（2n-1）×2ⁿ⁺¹-6，
∴Tn=6+(2n-1)×2n+1-2n+2．
則-Sn=6+(2n-1)×2n+1-2n+2-n2×2n+1=6-（n-1）²×2ⁿ⁺¹-2ⁿ⁺²．
∴Sn=(n-1)2×2n+1+2n+2-6．

點(diǎn)評：本題考查了錯(cuò)位相減法求數(shù)列的和，考查了學(xué)生的計(jì)算能力，是中檔題．