问题1.找到以下几何级数的总和:
(i)2、6、18,…至7个学期
解决方案:
Given G.P. has first term(a) = 2, common ratio(r) = 6/2 = 3 and number of terms(n) = 7
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
S7 = 2(37–1)/(3–1)
= 2(37–1)/2
= 2187–1
= 2186
Therefore, sum of 7 terms of the G.P. is 2186.
(ii)1、3、9、27,…至8个学期
解决方案:
Given G.P. has first term(a) = 1, common ratio(r) = 3 and number of terms(n) = 8
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
S8 = 1(38–1)/(3–1)
= 6560/2
= 3280
Therefore, sum of 8 terms of the G.P. is 3280.
(iii)1,–1/2,1/4,–1/8,…。
解决方案:
Given G.P. has first term(a) = 1, common ratio(r) = –1/2 and number of terms(n) is infinite.
We know sum of n terms of an infinite GP is given by S = a/(1–r).
S = 1/[1 – (–1/2)]
= 1/(3/2)
= 2/3
Therefore, sum of infinite terms of the G.P. is 2/3.
(iv)(a 2 – b 2 ),(a – b),(a–b)/(a + b),…至n个项
解决方案:
Given G.P. has first term(a) = (a2 – b2), common ratio(r) = (a – b)/(a2 – b2) = 1/(a+b) and number of terms is n.
We know sum of n terms of an infinite GP is given by Sn = a(rn–1)/(r–1).
Sn =
=
=
Therefore, sum of n terms of the G.P. is .
(v)4,2,1,1/2…至10个学期
解决方案:
Given G.P. has first term(a) = 4, common ratio(r) = 2/4 = 1/2 and number of terms(n) = 10.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
S10 =
=
=
=
Therefore, sum of 10 terms of the G.P. is .
问题2.找到以下几何级数的总和:
(i)0.15 + 0.015 + 0.0015 +…至8个词
解决方案:
Given G.P. has first term(a) = 0.15, common ratio(r) = 0.015/0.15 = 1/10 and number of terms(n) = 8.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
S8 =
= 0.15 (10/9) (1 – 1/108)
= (1/6) (1 – 1/108)
Therefore, sum of 8 terms of the G.P. is (1/6) (1 – 1/108).
(ii)√2+ 1 /√2+ 1 /2√2+…。至8个学期
解决方案:
Given G.P. has first term(a) = √2, common ratio(r) = (1/√2)/√2 = 1/2 and number of terms(n) = 8.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
S8 = √2[(1/2)8–1]/[(1/2)–1]
= √2(1–1/256)/(1/2)
= √2 (255/256) (2)
= (255√2)/128
Therefore, sum of 8 terms of the G.P. is (255√2)/128.
(iii)2/9 – 1/3 + 1/2 – 3/4 +…至5个学期
解决方案:
Given G.P. has first term(a) = 2/9, common ratio(r) = (–1/3)/(2/9) = –3/2 and number of terms(n) = 5.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
S5 =
=
=
Therefore, sum of 5 terms of the G.P. is .
(iv)(x + y)+(x 2 + xy + y 2 )+(x 3 + x 2 y + xy 2 + y 3 )+…。至n个词
解决方案:
Given series can be written as,
Sn = (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + . . . . to n terms
(x – y) Sn = (x + y) (x – y) + (x2 + xy + y2) (x – y) . . . to n terms
(x – y) Sn = x2 – y2 + x3 + x2y + xy2 – x2y – xy2 – y3 . . . to n terms
(x – y) Sn = (x2 + x3 + x4 + . . . n terms) + (y2 + y3 + y4 + . . . n terms)
(x – y) Sn = x2[(xn – 1)/(x – 1)] – y2[(yn – 1)/(y – 1)]
Sn = [x2[(xn – 1)/(x – 1)] – y2[(yn – 1)/(y – 1)]]/(x – y)
Therefore, sum of n terms of series is [x2[(xn – 1)/(x – 1)] – y2[(yn – 1)/(y – 1)]]/(x – y).
(v)3/5 + 4/5 2 + 3/5 3 + 4/5 4 +…至n个字词
解决方案:
Given series can be written as,
Sn = (3/5 + 3/53 + . . . to n terms) + (4/52 + 4/54 + . . . to n terms)
=
=
=
Therefore, sum of n terms of series is .
(六)
解决方案:
Given G.P. has first term(a) = , common ratio(r) = = and number of terms is n.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
Sn =
=
= –a i[1–(1+i)-n]
Therefore, sum of n terms of G.P. is –a i[1–(1+i)-n].
(vii)1 – –a,a 2 ,–a 3 ,…。 。 。 。至n项(a≠1)
解决方案:
Given G.P. has first term(a) = 1, common ratio(r) = –a and number of terms is n.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
Sn = [(–a)n–1]/(–a–1)
= [1–(–a)n]/(a+1)
Therefore, sum of n terms of G.P. is [1–(–a)n]/(a+1).
(viii)x 3 + x 5 + x 7 +。 。 。 。 n项
解决方案:
Given G.P. has first term(a) = x, common ratio(r) = x5/x3 = x2 and number of terms is n.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
= x3[x2n–1]/[x2–1]
Therefore, sum of n terms of G.P. is x3[x2n–1]/[x2–1].
(ix)√7+√21+3√7+。 。 。 。 n项
解决方案:
Given G.P. has first term(a) = √7, common ratio(r) = √21/√7 = √3 and number of terms = n.
We know sum of n terms of a GP is given by Sn = a(rn–1)/(r–1).
Sn = √7[(√3)n–1]/(√3–1)
Therefore, sum of n terms of G.P. is √7[(√3)n–1]/(√3–1).
问题3.评估以下内容:
(一世)
解决方案:
Given summation can be written as,
S11 = (2+31) + (2+32) + (2+33) + . . . . + (2+311)
= 2(11) + (31 + 32 + 33 + . . . . 311)
= 2(11) + 3(311–1)/(3–1)
= 22 + 265719
= 265741
Therefore, value of the summation is 265741.
(ii)
解决方案:
Given summation can be written as,
Sn = (2+30) + (22+31) + (23+32) + . . . . + (2n+3n-1)
= (21 + 22 + 23 + . . . . + 2n) + (30 + 31 + 32 + . . . . + 3n-1)
= 2(2n–1)/(2–1) + 30(3n–1)/(3–1)
= 2(2n–1) + (3n–1)/2
Therefore, value of the summation is 2(2n–1) + (3n–1)/2.
(iii)
解决方案:
Given summation can be written as,
S10-2+1 = S9 = 42 + 43 + 44 + . . . . 410
= 42(49–1)/(4–1)
= 16[49–1]/3
Therefore, value of the summation is 16[49–1]/3.
问题4.找到序列的总和:
(i)5 + 55 + 555 +…至n个词
解决方案:
We have Sn = 5 + 55 + 555 + ….. up to n terms.
Multiplying and dividing by 9, we get
= [9+99+999+…to n terms]
= [(10–1)+(102–1)+(103–1)…to n terms]
= [(10+102+103+….n terms) – (1+1+1+…..n terms)]
=
=
Therefore, the sum of the series up to n terms is .
(ii)7 + 77 + 777 +…至n个词
解决方案:
We have Sn = 7 + 77 + 777 + … to n terms.
Multiplying and dividing by 9, we get,
= [9+99+999+…to n terms]
= [(10–1)+(102–1)+(103–1)…to n terms]
= [(10+102+103+….n terms) – (1+1+1+…..n terms)]
=
=
Therefore, the sum of the series up to n terms is .
(iii)9 + 99 + 999 +…至n个词
解决方案:
We have Sn = 9 + 99 + 999 + … to n terms. It can be written as,
= (10–1)+(102–1)+(103–1)…to n terms
= (10+102+103+….n terms) – (1+1+1+…..n terms)
=
Therefore, the sum of the series up to n terms is .
(iv)0.5 + 0.55 + 0.555 +…至n项
解决方案:
We have Sn = 0.5 + 0.55 + 0.555 + … to n terms. It can be written as,
=
=
=
=
=
Therefore, the sum of the series up to n terms is .
(v)0.6 + 0.66 + 0.666 +…至n项
解决方案:
We have Sn = 0.6 + 0.66 + 0.666 + … to n terms. It can be written as,
=
=
=
=
=
Therefore, the sum of the series up to n terms is .
问题5.将GP 3、3 / 2、3 / 4,…的多少项加在一起即可得到3069/512?
解决方案:
Given G.P. has first term(a) = 3, common ratio(r) = (3/2)/3 = 1/2 and sum of terms(Sn) = 3069/512.
We know sum of n terms of a G.P. is given by Sn = a(rn–1)/(r–1).
=> 3069/512 = 3[1–(1/2)n] / [1–(1/2)]
=> 2(2n–1)/(2n) = 1023/512
=> 1023(2)n = 1024(2)n – 1024
=> 2n = 1024
=> n = 10
Therefore, 10 terms of the G.P. should be taken together to make 3069/512.
问题6:2 + 6 + 18 +…系列中的多少个术语。必须使总和等于728?
解决方案:
Given G.P. has first term(a) = 2, common ratio(r) = 6/2 = 3 and sum of terms(Sn) = 728.
We know sum of n terms of a G.P. is given by Sn = a(rn–1)/(r–1).
=> 728 = 2[3n–1]/[3–1]
=> 3n–1 = 728
=> 3n = 729
=> n = 6
Therefore, 6 terms of the G.P. must be taken together to make the sum equal to 728.
问题7.必须使用序列√3、3、3√3,…中的多少项才能使总和为39+13√3?
解决方案:
Given G.P. has first term(a) = 2, common ratio(r) = 3/√3 = 1/√3 and sum of terms(Sn) = 39+ 13√3.
We know sum of n terms of a G.P. is given by Sn = a(rn–1)/(r–1).
=> 39+13√3 = √3[3n/2–1]/(√3–1)
=> (39+13√3)(√3–1) = √3(3n/2–1)
=> 39√3–39+39–13√3 = 3(n+1)/2–√3
=> 3(n+1)/2 = 27√3
=> 3n/2 √3 = 27√3
=> 3n/2 = 27
=> n/2 = 3
=> n = 6
Therefore, 6 terms of the G.P. must be taken to make the sum 39+ 13√3.
问题8. GP 3、6、12,…的n个项的总和为381。求n的值。
解决方案:
Given G.P. has first term(a) = 3, common ratio(r) = 6/3 = 2 and sum of terms(Sn) = 381.
We know sum of n terms of a G.P. is given by Sn = a(rn–1)/(r–1).
=> 381 = 3(2n–1)/(2–1)
=> 2n – 1 = 127
=> 2n = 128
=> n = 7
Therefore, value of n is 7.
问题9. GP的公共比率是3,最后一项是486。如果这些项的总和是728,则找到第一个项。
解决方案:
Given G.P. has common ratio(r) = 3, last term(an) = 486 and sum of terms(Sn) = 728.
We know nth term of a G.P. is given by an = arn-1.
=> 486 = a(3)n-1
=> 486 = a(3)n/3
=> a(3)n = 1458 . . . . (1)
We know sum of n terms of a G.P. is given by Sn = a(rn–1)/(r–1).
=> 728 = a(3n–1)/(3–1)
=> 1456 = a(3)n–a
Using (1) in the equation, we get,
=> a = 1458 – 1456 = 2
Therefore, first term of the G.P. is 2.
问题10. GP的前三个项之和与前6个项之和的比率为125:152。找到共同的比率。
解决方案:
We know sum of n terms of a G.P. is given by Sn = a(rn–1)/(r–1).
According to the question, we have,
=>
=> (r3–1)/(r6–1) = 125/152
=> 1/(r3+1) = 125/152
=> 125r3 + 125 = 152
=> r3 = 27/125
=> r = 3/5
Therefore, the common ratio is 3/5.
问题11:GP的第4和第7项分别为1/27和1/729。找出GP的n个项的总和
解决方案:
We know nth term of a G.P. is given by an = arn-1.
According to the question, we have,
=> ar3 = 1/27 . . . . (1)
=> ar6 = 1/729 . . . . (2)
Dividing (2) by (1), we get,
=> r3 = 27/729 = 1/27
=> r = 1/3
Putting r = 1/3 in (1), we get,
=> a(1/3)3 = 1/27
=> a(1/27) = 1/27
=> a = 1
We know sum of n terms of a G.P. is given by Sn = a(rn–1)/(r–1).
=> Sn = 1[(1/3)n–1]/[(1/3)–1]
= 3[1–(1/3)n]/2
Therefore, sum of n terms of the G.P. is 3[1–(1/3)n]/2.