Sequence and Series Class 11 OP Malhotra Exe-14I ISC Maths Solutions Ch-14 Solutions. In this article you would learn about Series Involving Natural Numbers. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Sequence and Series Class 11 OP Malhotra Exe-14I ISC Maths Solutions Ch-14

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-14	Sequence and Series
Writer	O.P. Malhotra
Exe-14(I)	Series Involving Natural Numbers.

Exercise- 14I

Sequence and Series Class 11 OP Malhotra Exe-14I Solution.

Que-1: Find the sum to n terms of the series whose nth term is
(i) n(n + 2)
(ii) 3n² + 2n
(iii) 4n³ + 6n² + 2n

Sol: (i) Given T_n = n(n + 2)=n² + 2n
∴ S_n = ΣT_n = Σ (n² + 2n) = Σn² + 2Σn
= [{n(n+1)(2n+1)}/6] + [{2n(n+1)}/2]
= [{n(n+1)}/6] [2n + 1 + 6]
= {n(n+1)(2n+7)}/6

(ii) Given T_n = 3n² + 2n
∴ S_n = ΣT_n = 3 Σn² + 2Σn
= [{3n(n+1)(2n+1)}/6] + [{2n(n+1)}/2]
= [{n(n+1)}/2] [2n + 1 + 2]
= {n(n+1)(2n+3)}/2

(iii) Given T_n = 4n³ + 6n² + 2n
∴ S_n = ΣT_n = 4 Σn³ + 6Σn² + 2Σn
= [{4n²(n+1)²}/4] +[{6n(n+1)(2n+1)}/6] + [{2n(n+1)}/2]
= n(n + 1)[n(n + 1) + 2n + 1 + 1]
= n(n + 1)[n(n + 1) + 2(n + 1)]
= n(n + 1)²(n + 2)

Que-2: Find the sum of the series
(i) 3 × 5 + 5 × 7 + 7 × 9 + ….. to n terms
(ii) 1² + 3² + 5² +….. to n terms
(iii) 2² + 4² + 6² + ….. to n terms.

Sol: (i) T_n = (nth term of 3, 5, 7, ….)(nth term of 5, 7, 9, …)
= [3 + 2(n – 1)][5 + 2(n – 1)] = (2n + 1)(2n + 3) = 4n² + 8n + 3
∴ S_n = ΣT_n = 4Σn² + 8Σn + Σ3 = [{4n(n+1)(2n+1)}/6] + [{8n(n+1)}/2] + 3n
= (n/3) [2(n + 1)(2n + 1) + 12(n + 1) + 9] = (n/3) [4n² + 18n + 23]

(ii) Here Tn = [nth term of 1, 3, 5, ……]²
= [1 + 2(n – 1)]² = (2n – 1)² = 4n² – 4n + 1
∴ Sn = ΣTn = 4Σn² – 4Σn + n
= [{4n(n+1)(2n+1)}/6] – [{4n(n+1)}/2] + n
= [{2n(n+1)(2n+1)}/3] – 2n (n + 1) + n
= (n/3) [2(n + 1)(2n + 1) – 6(n + 1) + 3]
= (n/3) [4n² + 6n + 2 – 6n – 6 + 3] = (n/3) [4n² – 1]
= (n/3) (2n – 1) (2n + 1)

(iii) Tn = (nth term of 2, 4, 6,……..)² = [2 + (n – 1) 2]²
⇒ Tn = 4n²
∴ Sn = ΣTn = 4Σn²
= [{4n(n+1)(2n+1)}/6]
= {2n(n+1)(2n+1)}/3

Que-3: Find the nth term and the sum to n terms of the series 1.2 + 2.3 + 3.4 + ……

Sol: ∴ T_n = (nth term of 1, 2, 3, ……)(nth term of 2, 3, 4, …..)
= [1 + (n + 1) . 1] [2 + (n – 1) . 1] = n(n + 1) = n² + n
∴ S_n = ΣT_n = Σn² + Σn = [{n(n+1)(2n+1)}/6] + [{n(n+1)}/2]
= [{n(n+1)}/6] [2n + 1 + 3]
= [{n(n+1)(2n+4)}/6] = [{n(n+1)(n+2)}/3]

Que-4: Sum up to n terms the series 1.2² + 2.3² + 3.4² + …..

Sol: T_n = (nth term of 1, 2, 3, …..)(nth term of 2, 3, 4, …….)² = n(n + 1)² = n(n² + 2n + 1)
∴ T_n = n³ + 2n² + n

Que-5: Sum up 1 +(1 + 2) + (1 + 2 + 3) + ……. +(1 + 2 + 3 + ….. + n).

Sol: Here Tn = 1 + 2 + 3 + …… + n
= Σn = {n(n+1)}/2 = (1/2) [n² + n]
∴ Sn = ΣTn = (1/2) [Σn² + Σn]
= (1/2) [[{n(n+1)(2n+1)}/6] + [{n(n+1)}/2]]
= [{n(n+1)}/12] (2n+1+3) = [{n(n+1)(2n+4)}/12]
= {n(n+1)(n+2)}/6

Que-6: Sum up to n terms the series 1 + (1+(1/2)) + (1+(1/2)+(1/4)) ….

Sol: Tn = 1 + 1/2 + 1/4 + ……. n terms

Que-7: Sum up to n terms the series where nth term is 2ⁿ – 1.

Sol: Given T_n = 2ⁿ – 1
∴S_n = ΣT_n = Σ(2ⁿ – 1) = Σ2ⁿ – n = (2¹ + 2² + 2³ …… n terms) – n
= {2(2^n−1)}/(2−1) − n
= 2ⁿ⁺¹ – 2 – n

Que-8: Sum up 1 + 4 + 8 + 13 + ….. to n terms.

Sol: Let S_n = 1 + 4 + 8 + 13 +……. + T_n-1 + T_n …(1)
S_n = 1 + 4 + 8 + ….. + T_n-1 + T_n …(2)
0 = 1 + [3 + 4 + 5 + ….. (n – 1) terms] – T_n
T_n = 1 + {(n−1)/2} [6 + (n – 1 – 1) 1] = 1 + (6 + n – 2) {(n−1)/2}
⇒ Tn = 1 + [(n+4)(n−1)/2] = (1/2) [n^n + 3n + 2]
∴ Sn = ΣTn = (1/2) [Σn² + 3Σn – 2n]
= (1/2) [[{n(n+1)(2n+1)}/6] + [{3n(n+1)}/2] − 2n]
= (n/12) [(n + 1) (2n + 1) + 9 ( n + 1) – 12]
= (n/12) (2n² + 12n – 2) = (n/6) [n² + 6n – 1]

Que-9: Sum up 3 + 5 + 11 + 29 + ….. to n terms.

Sol: Let S = 3 + 5 + 11 + 29 ……. + T_n-1 + T_n …(1)
∴ S = 3 + 5 + 11 + …. + T_n-1 + T_n …(2)
eqn. (1) – eqn. (2) gives;
0 = 3 + [2 + 6 + 18 + ….. +(n – 1) terms] – T_n
⇒ T_n = 3 + [2{3^(n−1)−1}]/(3−1)
⇒ T_n = 3 + 3^n-1 – 1 = 3^n-1 + 2
∴ S_n = ΣT_n = Σ3^n-1 + 2n = [3 + 3¹ + 3² + …….. + 3^n-1] + 2n
= [1{3^n−1}/(3−1)] + 2n = [{1(3^n−1)}/2] + 2n

Que-10: Sum to n terms the series 7 + 77 + 777 + ……..

Sol: Let S_n = 7 + 77 + 777 + …. + T_n-1 + T_n …(3)
∴ S_n = 7 + 77 + ….. + T_n-1 + T_n …(2)
eqn. (1) – eqn. (2) gives ;
0 = 7 + [70 + 700 + …… (n – 1) terms ] – T_n
⇒ T_n = 7 + 70 + 700 …… n terms = {7[10^n−1]}/(10−1) = (7/9) [10^n – 1]
∴ Sn = ΣTn = (7/9) [Σ10^n – n]
= (7/9) [10 + 10² + 10³ … n terms – n]
= (7/9) [{10(10^n−1)}/(10−1) − n]
= (7/9) [(10/9) (10^n−1)−n]
= (70/81) (10^n−1) – (7n/9)

Que-11: Sum to n terms the series 1 + 3 + 7 + 15 + 31 + ….

Sol: Let S_n = 1 + 3 + 7 + 15 + 31 + ……. + T_n-1 + T_n …(1)
∴ S_n = 1 + 3 + 7 + 15 + …… +….. T_n-1 + T_n …(2)
eqn. (1) – eqn. (2) gives ;
0 = 1 + 2 + 4 + 8 + 16 + …… n terms – T_n
⇒ T_n = {1[2^n−1]}/(2−1) = 2^n
∴ S_n = ΣT_n = Σ(2ⁿ – 1) = Σ2ⁿ – n = (2¹ + 2² + … + n terms) – n
= [{2(2^n−1)}/(2−1)] – n = 2ⁿ⁺¹ – 2 – n

Que-12:  Find the sum to n terms of the series 1. 2. 3 + 2. 3. 4 + 3. 4. 5 + ……

Sol: T_n = (nth term of 1, 2, 3, …) × (nth term of 2, 3, 4, ……) × (nth term of 3, 4, 5, …)
= n(n + 1) (n + 2) = n (n² + 3n + 2) = n³ + 3n² + 2n
∴ S_n = ΣT_n = Σn³ + 3Σn² + 2Σn
= [{n²(n+1)²}/4] + [{3n(n+1)(2n+1)}/6] + [{2n(n+1)}/2]
= [{n(n+1)}/4] [n (n + 1) + 2 (2n + 1) + 4]
= [n² + 5n + 6] = {n(n+1)(n+2)(n+3)}/4

Que-13: Find the sum of the series to n terms and to infinity : 1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + …..

Sol:

Que-14: Sum to n terms the series whose nth term is 1/(4n²-1)

Sol: Tn = 1/(4n²-1) = 1/(2n-1)(2n+1)

Que-15: Natural numbers are written as

Show that the sum of the numbers in the nth group is (n/2) (n²+1)

Sol: First term of nth group is the nth term of series 1, 2, 4, …..
Let S_n = 1 + 2 + 4 + …… + T_n-1 + T_n …(1)
∴ S_n = 1 + 2 + …… + T_n-1 + T_n …(2)
∴ eqn. (1) – eqn. (2) gives ;
T_n = 1 + [1 + 2 + 3 ……(n – 1) terms]
∴ Tn = 1 + [{(n−1)n}/2] = {n²−n+2}/2
Thus first term of nth group = A = {n²−n+2}/2
and last term of nth group is the nth term of series 1, 3, 6, ……
Let S_n = 1 + 3 + 6 + 10 + …… + T_n-1 + T_n ….(3)
∴ S_n =1 + 3 + 6 + ….. + T_n-1 + T_n …(4)
eqn. (3) – eqn. (4) gives ;
0 = 1 + 2 + 3 + 4 + …… n terms – T_n
⇒ Tn = Σn = {n(n+1)}/2
∴ last term of nth group = L = {n(n+1)}/2
Hence required sum of numbers in nth groups = (n/2) [ A+L]
= (n/2) [{(n²−n+2)/2} + {(n(n+1))/2}]
= (n/4) [n²−n+2+n²+n]
= (n/4) [2n²+2] = (n/2) (n²+1)

Que-16: If the sum of first n terms of an A.P. is cn², then the sum of squares of these n terms is
(a) {n(4n²−1)c²}/6
(b) {n(4n²+1)c³}/3
(c) {n(4n²−1)c²}/3
(d) {n(4n²+1)c²}/6

Sol:

–: End of Sequence and Series Class 11 OP Malhotra Exe-14I ISC Maths Ch-14 Solutions. :–

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