OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Ex 14(i)

Continuous practice using Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Ex 14(i) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(i)

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

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

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

Question 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.
Solution:
(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 = \(\frac{4 n(n+1)(2 n+1)}{6}\) + \(\frac{8 n(n+1)}{2}\) + 3n
= \(\frac { n }{ 3 }\) [2(n + 1)(2n + 1) + 12(n + 1) + 9] = \(\frac { n }{ 3 }\) [4n² + 18n + 23]

(ii) Here T_n = [nth term of 1, 3, 5, ……]²
= [1 + 2(n – 1)]² = (2n – 1)² = 4n² – 4n + 1
∴ S_n = ΣT_n = 4Σn² – 4Σn + n = \(\frac{4 n(n+1)(2 n+1)}{6}\) – \(\frac{4 n(n+1)}{2}\) + n
= \(\frac{2 n(n+1)(2 n+1)}{3}\) – 2n (n + 1) + n
= \(\frac { n }{ 3 }\) [2(n + 1)(2n + 1) – 6(n + 1) + 3]
= \(\frac { n }{ 3 }\) [4n² + 6n + 2 – 6n – 6 + 3] = \(\frac { n }{ 3 }\) [4n² – 1]
= \(\frac { n }{ 3 }\) (2n – 1) (2n + 1)

(iii) T_n = (nth term of 2, 4, 6,……..)² = [2 + (n – 1) 2]²
⇒ T_n = 4n²
∴ S_n = ΣT_n = 4Σn² = \(\frac{4 n(n+1)(2 n+1)}{6}\) = \(\frac{2 n(n+1)(2 n+1)}{3}\)

Question 3.
Find the nth term and the sum to n terms of the series 1.2 + 2.3 + 3.4 + ……
Solution:
∴ 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 = \(\frac{n(n+1)(2 n+1)}{6}\) + \(\frac{n(n+1)}{2}\) = \(\frac{n(n+1)}{6}\)[2n + 1 + 3]
= \(\frac{n(n+1)(2 n+4)}{6}\) = \(\frac{n(n+1)(n+2)}{3}\)

Question 4.
Sum up to n terms the series 1.2² + 2.3² + 3.4² + …..
Solution:
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

Question 5.
Sum up 1 +(1 + 2) + (1 + 2 + 3) + ……. +(1 + 2 + 3 + ….. + n).
Solution:
Here T_n = 1 + 2 + 3 + …… + n = Σn = \(\frac{n(n+1)}{2}\) = \(\frac{1}{2}\) [n² + n]
∴ S_n = ΣT_n = \(\frac{1}{2}\) [Σn² + Σn] = \(\frac{1}{2}\) \(\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right]\)
= \(\frac{n(n+1)}{12}\)(2n+1+3) = \(\frac{n(n+1)(2 n+4)}{12}\) = \(\frac{n(n+1)(n+2)}{6}\)

Question 6.
Sum up to n terms the series 1 + \(\left(1+\frac{1}{2}\right)\) + \(\left(1+\frac{1}{2}+\frac{1}{4}\right)\) ….
Solution:
T_n = 1 + \(\frac{1}{2}\) + \(\frac{1}{4}\) + ……. n terms

Question 7.
Sum up to n terms the series where nth term is 2ⁿ – 1.
Solution:
Given T_n = 2ⁿ – 1
∴S_n = ΣT_n = Σ(2ⁿ – 1) = Σ2ⁿ – n = (2¹ + 2² + 2³ …… n terms) – n
= \(\frac{2\left(2^n-1\right)}{2-1}-n\)
\(\left[\text { Here } a=2 ; r=2>1 \text { and } \mathrm{S}_n=\frac{a\left(r^n-1\right)}{r-1}\right]\)
= 2ⁿ⁺¹ – 2 – n

Question 8.
Sum up 1 + 4 + 8 + 13 + ….. to n terms.
Solution:
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 + \(\left(\frac{n-1}{2}\right)\)[6 + (n – 1 – 1) 1] = 1 + (6 + n – 2)\(\left(\frac{n-1}{2}\right)\)
⇒ T_n = 1 + \(\frac{(n+4)(n-1)}{2}\) = \(\frac { 1 }{ 2 }\) [nⁿ + 3n + 2]
∴ S_n = ΣT_n = \(\frac { 1 }{ 2 }\) [Σn² + 3Σn – 2n] = \(\frac { 1 }{ 2 }\) \(\left[\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}-2 n\right]\)
= \(\frac { n }{ 12 }\) [(n + 1) (2n + 1) + 9 ( n + 1) – 12]
= \(\frac { n }{ 12 }\) (2n² + 12n – 2) = \(\frac { n }{ 6 }\) [n² + 6n – 1]

Question 9.
Sum up 3 + 5 + 11 + 29 + ….. to n terms.
Solution:
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 + \(\frac{2\left[3^{n-1}-1\right]}{3-1}\)
\(\left[\text { Here } a=2 \text { and } r=3>1 \text { and } \mathrm{S}_n=\frac{a\left(r^n-1\right)}{r-1}\right]\)
⇒ 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
= \(\frac{1\left[3^n-1\right]}{3-1}\) + 2n = \(\frac{1\left(3^n-1\right)}{2}\) + 2n

Question 10.
Sum to n terms the series 7 + 77 + 777 + ……..
Solution:
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 = \(\frac{7\left[10^n-1\right]}{10-1}\) = \(\frac{7}{9}\) [10ⁿ – 1]
∴ S_n = ΣT_n = \(\frac{7}{9}\) [Σ10ⁿ – n] = \(\frac{7}{9}\) [10 + 10² + 10³ … n terms – n]
= \(\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]\) = \(\frac{7}{9}\left[\frac{10}{9}\left(10^n-1\right)-n\right]\) = \(\frac{70}{81}\left(10^n-1\right)\) – \(\frac{7 n}{9}\)

Question 11.
Sum to n terms the series 1 + 3 + 7 + 15 + 31 + ….
Solution:
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 = \(\frac{1\left[2^n-1\right]}{2-1}\) = 2ⁿ
∴ S_n = ΣT_n = Σ(2ⁿ – 1) = Σ2ⁿ – n = (2¹ + 2² + … + n terms) – n
= \(\frac{2\left(2^n-1\right)}{2-1}\) – n = 2ⁿ⁺¹ – 2 – n

Question 12.
Find the sum to n terms of the series 1. 2. 3 + 2. 3. 4 + 3. 4. 5 + ……
Solution:
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
= \(\frac{n^2(n+1)^2}{4}\) + \(\frac{3 n(n+1)(2 n+1)}{6}\) + \(\frac{2 n(n+1)}{2}\) = \(\frac{n(n+1)}{4}\) [n (n + 1) + 2 (2n + 1) + 4]
= \(\) [n² + 5n + 6] = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Question 13.
Find the sum of the series to n terms and to infinity : \(\frac{1}{1.3}\) + \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + …..
Solution:

Question 14.
Sum to n terms the series whose nth term is \(\frac{1}{4 n^2-1}\)
Solution:

Question 15.
Natural numbers are written as

Show that the sum of the numbers in the nth group is \(\frac{n}{2}\)(n² + 1).
Solution:
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]
∴ T_n = 1 + \(\frac{(n-1) n}{2}\) = \(\frac{n^2-n+2}{2}\)
Thus first term of nth group = A = \(\frac{n^2-n+2}{2}\)
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
⇒ T_n = Σn = \(\frac{n(n+1)}{2}\)
∴ last term of nth group = L = \(\frac{n(n+1)}{2}\)
Hence required sum of numbers in nth groups = \(\frac{n}{2}[\mathrm{~A}+\mathrm{L}]\) = \(\frac{n}{2}\left[\frac{n^2-n+2}{2}+\frac{n(n+1)}{2}\right]\)
= \(\frac{n}{4}\left[n^2-n+2+n^2+n\right]\) = \(\frac{n}{4}\left[2 n^2+2\right]\) = \(\frac{n}{2}\left(n^2+1\right)\)

Question 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) \(\frac{n\left(4 n^2-1\right) c^2}{6}\)
(b) \(\frac{n\left(4 n^2+1\right) c^3}{3}\)
(c) \(\frac{n\left(4 n^2-1\right) c^2}{3}\)
(d) \(\frac{n\left(4 n^2+1\right) c^2}{6}\)
Solution: