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+....

Soln:

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

Sn=nk=1a1=nk=1k(k+1) =nk=1k2+nk=1k n(n+1)(2n+1)6+n(n+1)2 n(n+1)2(2n+13+1) n(n+1)2(2n+43) n(n+1)(n+2)3

 

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

Soln:

Given series = 1×2×3+2×3×4+3×4×5+...nthterm,an=n(n+1)(n+2)

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n

S0=nk=1ak =nk=1k3+3nk=1k2+2nk=1k =[n(n+1)2]2+3n(n+1)(2n+1)6+2n(n+1)2 =[n(n+1)2]2+n(n+1)(2n+1)2+n(n+1) =n(n+1)2[n(n+1)2+2n+1+2] =n(n+1)2[n2+n+4n+62] =n(n+1)4(n2+5n+6) =n(n+1)4(n2+2n+3n+6) =n(n+1)[n(n+2)+3(n+2)]4 =n(n+1)(n+2)(n+3)4

 

Q3. Find sum to nth term of the series 3×12+5×22+7×32+...

Soln:

Given series = 3×12+5×22+7×32+...nthterm,

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

Sn=nk=1ak =nk=1=(2k3+k2)=2nk=1k3+nk=1k2 =2[n(n+1)2]2+n(n+1)(2n+1)6 =n2(n+1)22+n(n+1)(2n+1)6 =n(n+1)2[n(n+1)+2n+13] =n(n+1)2[3n2+3n+2n+13] =n(n+1)(3n2+5n+1)6

 

Q4. Find sum to nth term of the series 11×2+12×3+13×4+....

Soln:

Given series = 11×2+12×3+13×4+....

nthterm,an=1n(n+1)=1n1n+1 a1=1112 a2=1213 a3=1314... an=1n1n+1

Column wise adding the above terms, we get

a1+a2+a3++an=[11+12+13+...+1n][12+13+14+...+1n+1+] Sn=11n+1=n+11n+1=nn+1

 

Q5. Find sum to nth term of the series 52+62+72+...+202nthterm,an=(n+4)2=n2+8n+16

Soln:

Sn=nk=1ak=nk=1(k2+8k+16) Sn=nk=1ak=8nk=1k+nk=116 =n(n+1)(2n+1)6+8n(n+1)2+16n

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

S16=16(16+1)(2×16+1)6+8×16×(16+1)2+16×16 =(16)(17)(33)6+(8)×16×(16+1)2+16×16 =(16)(17)(33)6+(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+

Soln:

Given series = 3×8+6×11+9×14+an

= (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) =nk=1k2+15nk=1k 9×n(n+1)(2n+1)6+15×n(n+1)2 =3n(n+1)(2n+1)2+15n(n+1)2 =3n(n+1)2(2n+1+5) =3n(n+1)2(2n+6) =3n(n+1)(n+3)

 

Q7. Find sum to nth term of the series 12+(12+22)+(12+22+32)+....

Soln:

Given series = 12+(12+22)+(12+22+32)+....an

=(12+22+32+....+n2) =n(n+1)(2n+1)6 =n(2n2+3n+1)6 =2n3+3n2+n)6 =13n3+12n2+16n Sn=nk=1ak =nk=1(13k3+12k2+16k) =13nk=1k3+12nk=1k2+16nk=1k =13n2(n+1)2(2)2+12×n(n+1)(2n+1)6+16×n(n+1)2 =n(n+1)6[n(n+1)2+2n+12+12] =n(n+1)6[n2+n+2n+1+12] =n(n+1)6[n2+n+2n+22] =n(n+1)6[n(n+1)+2(n+1)2] =n(n+1)6[(n+1)(n+2)2] =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 Sn=nk=1ak=nk=1k3+5nk=1k2+4nk=1k =n2(n+1)24+5n(n+1)(2n+1)6+4n(n+1)2 =n(n+1)2[n(n+1)2+5(2n+1)6+4] =n(n+1)2[3n2+3n+20n+10+246] =n(n+1)2[3n2+23n+346] n(n+1)(3n2+23n+34)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 ……………(1)

Take nk=12k=21+22+22+....

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) ……………… (2)

From (1) and (2), we get

Sn=nk=1k2+2(2n1)=n(n+1)(2n+1)6+2(2n1)

 

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) =4nk=1k24nk=1k+nk=11 =4n(n+1)(2n+1)64n(n+1)2+n =2n(n+1)(2n+1)32n(n+1)+n =n[2(2n2+3n+1)32(n+1)+1] =n[4n2+6n+26n6+33] =n[4n213] =n(2n+1)(2n1)3