Class 11 Maths Ncert Solutions Chapter 9 Ex 9.4 Sequences & Series PDF

Class 11 Maths Ncert Solutions Ex 9.4

Class 11 Maths Ncert Solutions Chapter 9 Ex 9.4

Q1. Find sum to nth term for the series 1×2+2×3+3×4+4×5+....$1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + . . . .$

Soln:

Given series = 1×2+2×3+3×4+4×5+....$1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + . . . .$ nth term, an = n (n + 1)

Sn=nk=1a1=nk=1k(k+1)$S_{n} = \sum_{k = 1}^{n} a_{1} = \sum_{k = 1}^{n} k ( k + 1)$ =nk=1k2+nk=1k$= \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k$ n(n+1)(2n+1)6+n(n+1)2$\frac{n (n + 1) (2n + 1)}{6} + \frac{n (n + 1)}{2}$ n(n+1)2(2n+13+1)$\frac{n (n + 1)}{2} \left ( \frac{ 2n + 1}{3} + 1 \right )$ n(n+1)2(2n+43)$\frac{n (n + 1)}{2} \left ( \frac{ 2n + 4}{3} \right )$ n(n+1)(n+2)3$\frac{n (n + 1) (n + 2) }{3}$

Q2. Find sum to nth term of the series 1×2×3+2×3×4+3×4×5+.....$1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + . . . . .$

Soln:

Given series = 1×2×3+2×3×4+3×4×5+...nthterm,an=n(n+1)(n+2)$1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + . . . n^{th} \; term, \; a_{n} = n (n + 1) (n + 2)$

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n

S0=nk=1ak$S_{0} = \sum_{k = 1}^{n} a_{k}$ =nk=1k3+3nk=1k2+2nk=1k$= \sum_{k = 1}^{n} k^{3} + 3 \sum_{k = 1}^{n} k^{2} + 2 \sum^{n}_{k = 1} k$ =[n(n+1)2]2+3n(n+1)(2n+1)6+2n(n+1)2$= \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{3n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2}$ =[n(n+1)2]2+n(n+1)(2n+1)2+n(n+1)$= \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{n(n + 1)(2n + 1)}{2} + n(n + 1)$ =n(n+1)2[n(n+1)2+2n+1+2]$= \frac{n(n + 1)}{2} \left [ \frac{n(n + 1)}{2} + 2n + 1 + 2 \right ]$ =n(n+1)2[n2+n+4n+62]$= \frac{n(n + 1)}{2} \left [ \frac{n^{2} + n + 4n + 6}{2} \right ]$ =n(n+1)4(n2+5n+6)$= \frac{n(n + 1)}{4} \left ( n^{2} + 5n + 6 \right )$ =n(n+1)4(n2+2n+3n+6)$= \frac{n(n + 1)}{4} \left ( n^{2} + 2n + 3n + 6 \right )$ =n(n+1)[n(n+2)+3(n+2)]4$= \frac{n(n + 1)\left [ n (n + 2) + 3 (n + 2) \right ]}{4}$ =n(n+1)(n+2)(n+3)4$= \frac{n(n + 1) (n + 2) (n + 3) }{4}$

Q3. Find sum to nth term of the series 3×12+5×22+7×32+...$3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + . . .$

Soln:

Given series = 3×12+5×22+7×32+...nthterm,$3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + . . . n^{th} term,$

an = (2n + 1) n2 = 2n3 + n2

Sn=nk=1ak$S_{n} = \sum^{n}_{k = 1} a_{k}$ =nk=1=(2k3+k2)=2nk=1k3+nk=1k2$= \sum^{n}_{k = 1} = (2k^{3} + k^{2}) = 2\sum^{n}_{k = 1} k^{3} + \sum^{n}_{k = 1} k^{2}$ =2[n(n+1)2]2+n(n+1)(2n+1)6$= 2 \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{n(n + 1)(2n + 1)}{6}$ =n2(n+1)22+n(n+1)(2n+1)6$= \frac{n^{2} (n + 1)^{2} }{2} + \frac{n(n + 1)(2n + 1)}{6}$ =n(n+1)2[n(n+1)+2n+13]$= \frac{n (n + 1) }{2} \left [ n(n + 1) + \frac{2n + 1}{3} \right ]$ =n(n+1)2[3n2+3n+2n+13]$= \frac{n (n + 1) }{2} \left [ \frac{3n^{2} + 3n + 2n + 1}{3} \right ]$ =n(n+1)(3n2+5n+1)6$= \frac{n (n + 1) \left ( 3n^{2} + 5n + 1 \right ) }{6}$

Q4. Find sum to nth term of the series 11×2+12×3+13×4+....$\frac{1}{ 1 \times 2 } + \frac{1}{ 2 \times 3 } + \frac{1}{ 3 \times 4 } + . . . .$

Soln:

Given series = 11×2+12×3+13×4+....$\frac{1}{ 1 \times 2 } + \frac{1}{ 2 \times 3 } + \frac{1}{ 3 \times 4 } + . . . .$

nthterm,an=1n(n+1)=1n1n+1$n^{th} \; term, a_{n} = \frac{1}{n(n + 1)} = \frac{1}{n} – \frac{1}{n + 1}$ a1=1112$a_{1} = \frac{1}{1} – \frac{1}{2}$ a2=1213$a_{2} = \frac{1}{2} – \frac{1}{3}$ a3=1314...$a_{3} = \frac{1}{3} – \frac{1}{4} . . .$ an=1n1n+1$a_{n} = \frac{1}{n} – \frac{1}{n + 1}$

Column wise adding the above terms, we get

a1+a2+a3++an=[11+12+13+...+1n][12+13+14+...+1n+1+]$a_{1} +a_{2} + a_{3} + … + a_{n} = \left [ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n} \right ] – \left [ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{n + 1} +\right ]$ Sn=11n+1=n+11n+1=nn+1$S_{n} = 1 – \frac{1}{n + 1} = \frac{n + 1 – 1}{n + 1} = \frac{n}{n + 1}$

Q5. Find sum to nth term of the series 52+62+72+...+202nthterm,an=(n+4)2=n2+8n+16$5^{2} + 6^{2} + 7^{2} + . . . + 20^{2} \; n^{th} \; term, \; a_{n} = (n + 4)^{2} = n^{2} + 8 n + 16$

Soln:

Sn=nk=1ak=nk=1(k2+8k+16)$S_{n} = \sum^{n}_{k = 1} a_{k} = \sum^{n}_{k = 1}(k^{2} + 8k + 16)$ Sn=nk=1ak=8nk=1k+nk=116$S_{n} = \sum^{n}_{k = 1} a_{k} = 8\sum^{n}_{k = 1}k + \sum^{n}_{k = 1}16$ =n(n+1)(2n+1)6+8n(n+1)2+16n$= \frac{n(n + 1)(2n + 1)}{6} + \frac{8n (n + 1)}{2} + 16 n$

16th term is (16 + 4)2 = 202

S16=16(16+1)(2×16+1)6+8×16×(16+1)2+16×16$S_{16} = \frac{16(16 + 1)(2 \times 16 + 1)}{6} + \frac{8 \times 16 \times (16 + 1)}{2} + 16 \times 16$ =(16)(17)(33)6+(8)×16×(16+1)2+16×16$= \frac{(16)(17)(33)}{6} + \frac{(8) \times 16 \times (16 + 1)}{2} + 16 \times 16$ =(16)(17)(33)6+(8)(16)(17)2+256$= \frac{(16)(17)(33)}{6} + \frac{(8) (16) (17)}{2} + 256$

= 1496 + 1088 + 256

= 2840

52 + 62 + 72 + …. + 202 = 2840

Q6. Find sum to nth term of the series 3×8+6×11+9×14+$3 \times 8 + 6 \times 11 + 9 \times 14 + …$

Soln:

Given series = 3×8+6×11+9×14+an$3 \times 8 + 6 \times 11 + 9 \times 14 + … a_{n}$

= (nth term of 3, 6, 9, …) x (nth term of 8, 11, 14 .. .)

= (3n) (3n + 5)

= 9n2 + 15n

Sn=nk=1ak=nk=1(9k2+15k)$S_{n} = \sum^{n}_{k = 1}a_{k} = \sum^{n}_{k = 1} (9k^{2} + 15k)$ =nk=1k2+15nk=1k$= \sum^{n}_{k = 1}k^{2} + 15 \sum^{n}_{k = 1} k$ 9×n(n+1)(2n+1)6+15×n(n+1)2$9 \times \frac{n(n + 1)(2n + 1)}{6} + 15 \times \frac{n (n + 1)}{ 2}$ =3n(n+1)(2n+1)2+15n(n+1)2$= \frac{3n (n + 1)(2n + 1)}{2} + \frac{ 15n (n + 1)}{ 2}$ =3n(n+1)2(2n+1+5)$= \frac{3n (n + 1)}{2}(2n + 1 + 5)$ =3n(n+1)2(2n+6)$= \frac{3n (n + 1)}{2}(2n + 6)$ =3n(n+1)(n+3)$= 3n (n + 1)(n + 3)$

Q7. Find sum to nth term of the series 12+(12+22)+(12+22+32)+....$1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . .$

Soln:

Given series = 12+(12+22)+(12+22+32)+....an$1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . . a^{n}$

=(12+22+32+....+n2)$= \left (1^{2} + 2^{2} + 3^{2} + . . . . + n^{2} \right )$ =n(n+1)(2n+1)6$= \frac{n(n + 1)(2n + 1)}{6}$ =n(2n2+3n+1)6$= \frac{n(2n^{2} + 3n + 1)}{6}$ =2n3+3n2+n)6$= \frac{2n^{3} + 3n^{2} + n)}{6}$ =13n3+12n2+16n$= \frac{1}{3}n^{3} + \frac{1}{2}n^{2} + \frac{1}{6}n$ Sn=nk=1ak$S_{n} = \sum_{k = 1}^{n} a_{k}$ =nk=1(13k3+12k2+16k)$= \sum_{k = 1}^{n} \left ( \frac{1}{3} k^{3} + \frac{1}{2} k^{2} + \frac{1}{6} k \right )$ =13nk=1k3+12nk=1k2+16nk=1k$= \frac{1}{3} \sum_{k = 1}^{n} k^{3} + \frac{1}{2} \sum_{k = 1}^{n} k^{2} + \frac{1}{6} \sum_{k = 1}^{n} k$ =13n2(n+1)2(2)2+12×n(n+1)(2n+1)6+16×n(n+1)2$= \frac{1}{3} \frac{n^{2}(n + 1)^{2}}{(2)^{2}} + \frac{1}{2} \times \frac{n(n + 1)(2n + 1)}{6} + \frac{1}{6} \times \frac{n(n + 1)}{2}$ =n(n+1)6[n(n+1)2+2n+12+12]$= \frac{n(n + 1)}{6} \left [ \frac{n(n + 1)}{2} + \frac{2n + 1}{2} + \frac{1}{2} \right ]$ =n(n+1)6[n2+n+2n+1+12]$= \frac{n(n + 1)}{6} \left [ \frac{n^{2} + n + 2n + 1 + 1 }{2} \right ]$ =n(n+1)6[n2+n+2n+22]$= \frac{n(n + 1)}{6} \left [ \frac{n^{2} + n + 2n + 2 }{2} \right ]$ =n(n+1)6[n(n+1)+2(n+1)2]$= \frac{n(n + 1)}{6} \left [ \frac{n( n + 1) + 2(n + 1) }{2} \right ]$ =n(n+1)6[(n+1)(n+2)2]$= \frac{n(n + 1)}{6} \left [ \frac{( n + 1) (n + 2) }{2} \right ]$ =n(n+1)2(n+2)12$= \frac{n(n + 1)^{2} (n + 2) }{12}$

Q8. Find sum to nth term of the series that has nth term given by n(n + 1)(n + 4)

Soln:

an=n(n+1)(n+4)=n(n2+5n+4)=n3+5n2+4n$a_{n} = n(n + 1)(n + 4) = n(n^{2} + 5n + 4) = n^{3} + 5n^{2} + 4n$ Sn=nk=1ak=nk=1k3+5nk=1k2+4nk=1k$S_{n} = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} k^{3} + 5 \sum_{k = 1}^{n} k^{2} + 4 \sum_{k = 1}^{n} k$ =n2(n+1)24+5n(n+1)(2n+1)6+4n(n+1)2$= \frac{n^{2} (n + 1)^{2}}{4} + \frac{5n (n + 1)(2n + 1)}{6} + \frac{4n( n + 1)}{2}$ =n(n+1)2[n(n+1)2+5(2n+1)6+4]$= \frac{n(n + 1)}{2} \left [ \frac{ n(n + 1)}{2} + \frac{5(2n + 1)}{6} + 4 \right ]$ =n(n+1)2[3n2+3n+20n+10+246]$= \frac{n(n + 1)}{2} \left [ \frac{ 3n^{2} + 3n + 20n + 10 + 24 }{6} \right ]$ =n(n+1)2[3n2+23n+346]$= \frac{n(n + 1)}{2} \left [ \frac{ 3n^{2} + 23n + 34 }{6} \right ]$ n(n+1)(3n2+23n+34)12$\frac{n(n + 1) \left ( 3n^{2} + 23n + 34 \right ) }{12}$

Q9. Find sum to nth term of the series that has nth term given by n2 + 2n

Soln:
an = n2 + 2n

Sn=nk=1k2+2k=nk=1k2+nk=12k$S_{n} = \sum_{k = 1}^{n} k^{2} + 2^{k} = \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} 2^{k}$ ……………(1)

Take nk=12k=21+22+22+....$\sum_{k = 1}^{n} 2^{k} = 2^{1} + 2^{2} + 2^{2} + . . . .$

The series 2, 22, 23, …. Is a G.P with 2 as first term as well as common ratio.

nk=12k=(2)[(2)n1]21=2(2n1)$\sum_{k = 1}^{n} 2^{k} = \frac{(2)\left [ (2)^{n} – 1 \right ]}{2 – 1} = 2 (2^{n} – 1)$ ……………… (2)

From (1) and (2), we get

Sn=nk=1k2+2(2n1)=n(n+1)(2n+1)6+2(2n1)$S_{n} = \sum_{k = 1}^{n} k^{2} + 2(2^{n} – 1) = \frac{n (n + 1)(2n + 1)}{6} + 2 (2^{n} – 1)$

Q10. Find sum to nth term of the series that has nth term given by (2n – 1)2

Soln:

an = (2n – 1)2 = 4n2 – 4n + 1

Sn=4nk=1ak4nk=1(4k24k+1)$S_{n} = 4 \sum_{k = 1}^{n} a_{k} – 4 \sum_{k = 1}^{n} (4k^{2} – 4k + 1)$ =4nk=1k24nk=1k+nk=11$= 4\sum_{k = 1}^{n} k^{2} – 4 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1$ =4n(n+1)(2n+1)64n(n+1)2+n$= \frac{4n (n + 1) (2n + 1)}{6} – \frac{4n (n + 1)}{2} + n$ =2n(n+1)(2n+1)32n(n+1)+n$= \frac{2n (n + 1) (2n + 1)}{3} – 2n (n + 1) + n$ =n[2(2n2+3n+1)32(n+1)+1]$= n \left [ \frac{2 (2n^{2} + 3n + 1)} {3} – 2 (n + 1) + 1 \right ]$ =n[4n2+6n+26n6+33]$= n \left [ \frac{4n^{2} + 6n + 2 – 6n – 6 + 3} {3} \right ]$ =n[4n213]$= n \left [ \frac{4n^{2} – 1} {3} \right ]$ =n(2n+1)(2n1)3$= \frac{n (2n + 1) (2n – 1)}{3}$