“數(shù)列的求和”例題解析

1、“數(shù)列的求和”例題解析例1.求下列數(shù)列的前n項(xiàng)和Sn:(1)1'2，2=，t，(n1-3(3)1 ,423311+ 2，81 2 17 ,78331 11+ 21解(1)Sn=12438Sn1尹)，2a6''(n=(1 + 2+3+ n) + Gn(n +1)2n(n=1+亠1 10歹)11 21) 1-歹3 :1(一 + '32_尹1匚+332131(158(1丄)32n)1321藥)(3)先對(duì)通項(xiàng)求和an = 1- Sn=(2 + 2 + + 2)1歹)141尹1+41尹2 2)+ (盲 + T + + )S234丄)32n)132(1 +32n，12門(mén)1_

2、2_32n)1尹1 1-+ 一 + + n 12421),1 1=2n- (1 + - + + + 241=2n- 2 +12n-1)例2.求和:1 1+ 2 313 71581 2112 51+3 415918 11 1n(n 1)1(2n 1)(2 n 3) J(3n2)(1)n(n + 1)1 1n.1- Sn(-)(1221n 1丄)3(14)(-n11 n 1nn 1(2)(2 n 1)(2 n+3)111二 Sn =-1- n 八5 3712n 1412n 11(4 2n 11 1 15912n 32n1二)12n1=411 13 2n 1n(4n 5)3(2 n 1)(2 n 3

3、)12n3(3)'氣3 n 1)(3 n+2)11)13'3n1 1(-)'58,1(813n 2)117)1(3n 113n 2)1 1二 _ (_3 2n6n 413n 2)例3. 求下面數(shù)列的前n項(xiàng)和:1 1 11+1, a+4，界+7，盯 +(3n-2)分析：將數(shù)列中的每一項(xiàng)拆成兩個(gè)數(shù)，一個(gè)數(shù)組成以11為公比的等比數(shù)列，另一個(gè)數(shù)a組成以3n- 2為通項(xiàng)的等差數(shù)列，分別求和后再合并。解設(shè)數(shù)列的通項(xiàng)為an，前n項(xiàng)和為Sn+ (3n2)- Sn = (1當(dāng)a = 1時(shí),Sn1(3n2) n23n2 n2說(shuō)明1n a11 - a等比數(shù)列的求和問(wèn)題,Sn(1 3n 2)n

4、nAa1nn1a a(3n1)n分 q=1與q豐1兩種情況討論?！纠?】設(shè)ak = 12 + 22 + k2(k N*)，則數(shù)列a1a2,的前n項(xiàng)之和是a36nn 13nI 16(n 1)n6(nn 21)設(shè)數(shù)列a11a2a3的通項(xiàng)為bn .an又 an = 12 + 22 + n21二一 n(n+ 1)(2 n + 1)66 1 1bn = = 6()n n(n +1) n n +1數(shù)列bn的前n項(xiàng)和Sn=b1 + b2+ bn=6(1=6(11 12 31n)6nn + 1例5.求在區(qū)間a,b (b>a,a,b N ）上分母是3的不可約分?jǐn)?shù)之和。3a 3a 1 3a 2b上分母為3的

5、所有分?jǐn)?shù)是詩(shī),3a 4 3a 53b 2a+ 1, a+ 2，b 1,3a1為首項(xiàng)，以-為公差的等差數(shù)列.33解法一區(qū)間a,1項(xiàng)數(shù)為 3b 3a+ 1,其和 S = 2 (3b 3a+ 1)(a + b)其中，可約分?jǐn)?shù)是 a, a + 1, a+ 2,，b1其和 S'= -(b a+ 1)(a + b)故不可約分?jǐn)?shù)之和為S S'= 1(a + b)(3b 3a+ 1) (b a+ 1)=b2 - a2解法二 S =3a +1 3a + 2 3a + 4 3a + 53b 2 3b 1+ +33333124521 S=(a + -) + (a + E)+(a+ 3) + (a +

6、 3) + + (b -)+ (b-)七12452而又有 S=(b -) + (b -)+ (b -) + (b -) + + (a + -) +1(a+ 3)兩式相加：2S= (a + b) + ( a+ b)+( a+ b)其個(gè)數(shù)為以3為分母的分?jǐn)?shù)個(gè)數(shù)減去可約分?jǐn)?shù)個(gè)數(shù)。即 3 (b a) + 1 ( b a+ 1) =2 (b a)2S=2 (b a) (a + b)S=b2- a2例6.求下列數(shù)列的前n項(xiàng)和Sn：(1)a,2a2, 3a3,，nan,，(am 0、1)；1,4, 9,，n2，;1,3x, 5x2,，(2n 1) xn-1,，(xm 1)234, 8解 (1) Sn=a +

7、 2a2+ 3a3 + nanaM 0 aSn=a2 + 2a3 + 3a4+( n 1) an+ nan+1Sn aSn=a+ a2+ a3+ an nan+1aM 1Sn亂1 an) nan 1 a)Snna1 aa(1 a ) na2(1 a) 1 aSn=1 + 4+ 9 + + n2(a+ 1) 3 a3=3a2 + 3a+123 13=3 X 12+ 3X 1+ 133 23=3X 22+ 3X 2+ 143 33=3 X 32 + 3 X 3+ 1n3 (n 1) 3=3 (n 1) 2+3 (n 1)+ 1(n+ 1) 3 n3=3n2 + 3n + 1把上列幾個(gè)等式的左右兩邊

8、分別相加，得(n + 1)3 13=3( 12+22+ n2)+ 3( 1 + 2+ n) + n2222 3n(n 1)=3(12 + 22 + 32 + + n2)+- +n12+ 22+ 32+ n213 3n(n 1)二-(n + 1)3 13213, C 2, C3n(n 1)二一 n + 3n + 3n32nn1 2=6n (2 n +3n+ 1)1二一n(n+ 1)(2n + 1)6(3)Sn=1 + 3x +5/+7乂3+( 2n 1) xn-1二xSn=x + 3x2 + 5x3 +( 2n 3) xn-1 +( 2n 1) xn兩式相減，得(1 x) Sn=1 + 2x (1 + x+ x2 + + xn-2) ( 2n 1) xn=1- (2n - 1)xn + 2x(xn1 1)(2n1)xn+1(2n1)xn (1 x)n+11 x(2n1)xn+1(2n1)xn (1 x)(1 x)2