若数An=(dn+e)×3ⁿ+(fn²+gn+h)×2ⁿ,{An}的前5项都等于76,求{An}的通

自立奉献 |浏览1015次
收藏|2019/02/04 13:09

满意回答

2019/02/04 13:16

A1 = ( d + e ) * 3 + ( f + g + h ) * 2 = 3d + 3e + 2f + 2g + 2h = 76;①A2 = ( 2d + e ) * 9 + ( 4f + 2g + h ) * 4 = 18d + 9e + 16f + 8g + 4h = 76;②A3 = ( 3d + e ) * 27 + ( 9f + 3g + h ) * 8 = 81d + 27e + 72f + 24g + 8h = 76;③A4 = ( 4d + e ) * 81 + ( 16f + 4g + h ) * 16 = 324d + 81e + 256f + 64g + 16h = 76;④A5 = ( 5d + e ) * 243 + ( 25f + 5g + h ) * 32 = 1215d + 243e + 800f + 160g + 32h = 76;⑤这是五元一次方程组,不需要用待定系数法。下面按要求不用行列式,只按普通方法消元。② -① * 2,12d + 3e + 12f + 4g = -76;⑥③ -① * 4,69d + 15e + 64f + 16g = -76 * 3;⑦④ -① * 8,300d + 57e + 240f + 48g = -76 * 7;⑧⑤ -① * 16,1167d + 195e + 768f + 128g = -76 * 15;⑨这是四元一次方程组,继续消元。⑦ -⑥ * 4,21d + 3e + 16f = 76;⑽⑧ -⑥ * 12,156d + 21e + 96f = 76 * 5;⑾⑨ -⑥ * 32,783d + 99e + 384f = 76 * 17;⑿这是三元一次方程组,继续消元。⑾ -⑽ * 6,30d + 3e = -76;⒀⑿ -⑽ * 24,279d + 27e = -76 * 7;⒁这是二元一次方程组,继续消元。⒁ -⒀ * 9,9d = 152,d = 152/9;代入⒀3e = -76 - 30d = -76 - 1520/3 = -1748/3,e = -1748/9;代入⑽16f = 76 - 21d - 3e = 76 - 21 *152/9 +1748/3 = 304,f = 19;代入⑥4g = -76 - 12d - 3e - 12f = -76 - 12 *152/9 +1748/3 - 12 * 19 = 76,g = 19;代入①2h = 76 -3d - 3e - 2f - 2g = 76 - 3 *152/9 +1748/3 - 2 * 19 - 2 * 19 = 532,h = 266;An 的通项公式 An = 76[ (2n- 23) * 3^(n-2) + ( n^2 + n + 14 ) * 2^(n-2) ] 。

寂园晓月

其他回答(0)
0人关注该问题
+1

 加载中...